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CH-3 (CURRENT ELECTRICITY)Session-11. Current Electricity: Study of electric charges in motion is called current electricity.2. Electric current: Charge flowing through any cross section of the conductor in one second.If a charge Q passes through an area in time t, then current I is given by, I = .If current is steady, i.e., rate of flow of charge does not change with time, then I=Q/t (rate of flow of charges); I= dq/dt (if rate of flow of charges varies with time).SI unit: ampere (A); 1A = 1C/1s. Electric current is a scalar quantity. Though, electric current has both magnitude and direction, but the laws of vector addition are not applicable to the addition of the electric currents. Thus, it is a scalar quantity. 1 milliampere = 1mA = 10-3A

1 microampere = 1A = 10-6A

Current in a domestic appliance = 1A

Current carried by a lightning = 104A

Types of current: (i)Alternating current (AC): It is a type of electriccurrent, in which the direction of the flow of electrons switches back and forthat regular intervals or cycles. Current flowing in power lines and normal household electricity that comes from a wall outlet is alternating current.(ii)Direct current (DC): It is electrical current which flows consistently in one direction. The current that flows in a flashlightor another appliance running on batteriesis direct current.Advantage of AC over DC is that it is relatively cheap to change the voltageof the current. Furthermore, the inevitable loss of energy that occurs when current is carried over long distances is far smaller with alternating current than with direct current. ( to be studied in session 6).3. Conventional current: The current whose direction is along the direction of the motion of positive charges under the action of electric field.The direction of electronic current is opposite to that of the conventional current. Electronic current is in the direction of flow of electrons. 4. Free charge carriers: The charged particles which by flowing in a definite direction set up an electric current are called current carriers. Different types of current carriers:a. Metals: In metallic conductors, electrons are charge carriers b. Semiconductors: In semiconductors both holes (positive charges) and electrons are charge carriers.c. Electrolytes and Ionised gases: Positively and negatively charged ions and electrons are charge carriers.

5. Ohms law: 1. The current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged.

2. V I V=IR, R is called proportionality constant called resistance. SI unit is ohm () and [ML2T-3A-2].

3. Any material that has some resistance is called a resistor.

4. R depends on nature of the conductor, its dimensions and physical conditions like temperature, etc.

5. The slope of V-I graph is resistance (R) of the given material. Higher the slope, more is the resistance on conductor.

6. Also, R l and R , hence R => R = , where is called resistivity or specific resistance. It depends upon nature of material of conductor, temperature and pressure. Its SI unit is ohm metre (m) and [ML3T-3A-2]. Graph for resistance v/s diameter is given beside (R).

7. Resistivity of a material is numerically equal to resistance of a conductor, having unit length and unit cross-section area.

Order of Resistivity () of different materials

Metals: 10-8m Semiconductors: 0.001 3000 mInsulators: 105 1016 m

6. Conductance: It is equal to reciprocal of resistance and is denoted by G. G = . SI unit is -1 or mho or siemens (S), Dimensional Formula= [M-1L-2T3A2].Conductivity: It is the reciprocal of resistvity and is denoted by . Its SI unit is -1m-1 or mho m-1 or Sm-1, Dimensional Formula = [M-1L-3T3A2]. 7. Cause of Resistance: When a potential difference is applied across a conductor, its free electrons get accelerated. On their way, they frequently collide with the positive metal ions and this opposition to the flow of electrons is called resistance. The number of collisions that the electrons make with the atoms/ions depends on the arrangement of atoms or ions in a conductor. With increase in temperature, metallic ions/atoms begin to vibrate more vigorously causing more opposition to flow of electrons and hence resistance of the material increases.8. Current Density: It is defined as the amount of charge flowing per second (Current) through a unit area. It is a vector quantity and is denoted by . j = . Its direction is along the area vector direction and its SI unit is Am-2, dimensions [AL-2].

9. Important Numerical Tools:i. For a wire of length l, cross-sectional area A and resistance R, if its length is increased by 10%, then its new length, l becomes 1.1l. By stretching a wire, its length increases, but its volume remains the same. Hence, Al= Al A =(Al)/(1.1l) =(A/1.1) Thus, R= = = (1.1)2R, where R= . Hence change in R is given by = (1.12 1)R = 0.21R = 21% increase.ii. Similarly if cross-section area changes, then we can apply the same steps as discussed in above point.iii. If charge (q) is given as a variable of t (time), then to find current, differentiate q w.r.t. t (time) since I=dq/dt. 10. If E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its ends is El, i.e., V = El Now, for a given conductor the resistance is directly proportional to the length of the conductor and inversely proportional to the area of cross-section, i.e.,R Hence, R = , where is the constant of proportionality. From, Ohms law we can write: V = IR = [From ] El = l [From ] E = [ = ], where is the current density.As the direction of current density is same as that of , = or = or = [ = 1/], where is known as conductivity. Vector Form of Ohms Law:

11. Expression for Drift Velocity:

Consider that , , .... , are the random velocities of N free electrons, then average velocity of the electrons will be = = 0 Thus, the average velocity, is zero since the directions of electrons after collision with the fixed ions is random.In the presence of , each e- experiences a force, -e and undergoes acceleration , i.e., = = , where m is the mass of an electron.Now, between two successive collisions, an e- gains a velocity component, randomly thermal velocity in opposite direction to .If an e- having random thermal velocity accelerates for time t1 (before it suffers collision) then, = +t1Similarly, the velocities of other e- will be = +t2 ........ = +tN

So the average velocity of all the N e-s is, = = + [From ] = 0 + , where = is the average time between two collision, i.e., relaxation time.Velocity gained by e- during this time is:

= = [From ]

12. Drift Velocity: It is defined as the average velocity gained by the free electrons of a conductor in the opposite direction of the externally applied electric field.When electrons drift in a metal from lower to higher potential, then all the electrons do not mean in the same direction. Random motion of electrons takes place.Relaxation Time: The average time that elapses between two successive collisions of an electron. On increasing temperature, its thermal motion increases, thus relaxation time decreases. Homework Sheet 1Total marks: 34 LEVEL 1 (15 marks)Theory (13 marks)1. Define electric current and give its SI unit.[1 mark]2. State Ohms law and its graph.[1 mark]3. Define drift velocity. A conductor of length L is connected to a dc source of emf E. If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L, how will the drift velocity change? [Delhi 2011,2marks]4. Define relaxation time of free electrons drifting in a conductor. How it is related to the drift velocity of electrons?[KV 2012, 2 marks]5. Define electrical resistivity and electrical conductivity of a material and give their SI unit. On what factors does the electrical resistivity of a material depend upon?[2 marks]6. What is the cause of resistance in a conductor?[2 marks]7. Define the term 'resistivity' and 'conductivity' and state their S.I. unit. [Delhi 2013, Outside Delhi 2012, Foreign 2011, 2marks]Application Based (2 marks)8. If a wire is stretched to double its length, without loss of mass how much will the resistivity of the wire change? Also, how much will the resistance change? [2 marks]LEVEL 2 (19 marks)Derivations (3 marks)1. *Is current a vector or scalar quantity? Deduce the expression for drift velocity of electrons in a conductor.(Refer Point 11) [DPS, 3 marks]Application Based (5 marks)2. *A potential difference of V volts is applied to a conductor of length L and diameter D. How are the electric field and the resistance of the conductor affected when in turn (i)V is halved, (ii) L is doubled and (iii) D is halved, where, in each case, the other two factors remain the same. Justify your answer in each case. [Foreign 2013,Outside Delhi 2012, 3marks]3. *Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker? [Outside Delhi 2012, Udgam, KV 2013, 1mark]4. When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? [Delhi 2012, Outside Delhi 2012, 1mark]Graph Based (1 mark)5. Draw a graph to show variation of resistance of a metal wire as a function of its diameter, keeping length and temperature constant. [1 mark]Numerical (10 marks)6. The graph given below shows how the current (I) varies with applied potential difference (v) across a 12 V, filament lamp (A) and across one metre long nichrome wire (B). Using the graph, find the ratio of values of the resistance of the filament lamp to the nichrome wire (i) when potential difference across them is 12V, (ii) when potential difference across them is 4V. How does the resistance of the filament lamp change as the current rises? Suggest a physical significance for this change. [Foreign 2013, 2 marks] Ans. (i) 3:4, (ii) 1:2. 7. A rheostat has 100 turns of a wire of radius 0.4 mm and having resistivity 4.210-7m. The diameter of each turn is 3 cm. What is the maximum value of resistance it can introduce? [2 marks]Ans. 7.8758. *A wire of resistance R=4 is stretched, so that its radius becomes 1/5th of its original value, calculate its new resistance.[MAV, 1 mark]Ans. 2.5k9. *The charge flowing in a conductor varies with time as, q=2t-6t2+10t3. Find initial current. [2 marks]Ans. 2A.10. A uniform wire of length l and radius r has resistance 100. It is recast into a thin wire of (i) length 2l (ii) radius r/2. Calculate the resistance of new wire in each case. [2 marks]Ans. 400, 1600.11. If length of a conducting wire is increased by 1% due to stretching, then by what percentage will the resistance change?[KV, 1mark]Ans. 2%.Session-21. Molecular concept of Resistivity:

Suppose a potential difference V is applied across a conductor of length l and of uniform cross-section A.The electric field set up inside the conductor is given by E = .Under the influence of field , the free electrons begin to drift in the opposite direction with an average velocity vd.Let n be the number of electrons per unit volume (number density).Thus, number of electrons in length l of conductor = n Volume of conductor = nAlTotal charge contained in length l of conductor, q = enAl Time taken by electrons to travel length of conductor, t = Current density, j = = nevd

Thus, Current I = = = neAvd

From vector form of Ohms law,E = j E = nevd = [From vd expression] Thus, =

2. Factors affecting drift velocity:i. With increase in temperature, drift velocity of electrons decrease, since resistance of the conductor increases and current is inversely proportional to resistance.ii. Drift velocity is inversely proportional to length, because on increasing length, resistance increases and hence current decreases. Thus, drift velocity decreases.iii. Drift velocity is proportional to area, because on increasing area, resistance decreases and hence current increases. Thus, drift velocity increases.3. Cause of instantaneous current: Although the drift speed of electrons is very small, of the order of 1mm/s, yet current is transferred very fast because electrons are present everywhere in an electric circuit (number density is high). When a potential difference is applied to a circuit, an electric field is set up throughout the circuit, with speed of light. Electrons in every part begin to drift under the influence of this electric field and a current begins to flow in the circuit almost immediately.So, though electron drift speed and electronic charge is very small, but still as number density is high, hence current through conductor is high.4. Mobility is defined as the magnitude of the drift velocity per unit electric field,. = As drift velocity,vd = For an electron, = , where m is the mass of an e-.SI unit of mobility is m2V-1s-1, dimensional formula = [M-1T2A].For Knowledge:Conductivityis proportional to the product of mobility and carrier concentration. For semiconductors, the behaviour oftransistorsand other devices can be very different depending on whether there are many electrons with low mobility or few electrons with high mobility. Therefore mobility is a very important parameter for semiconductor materials. Almost always, higher mobility leads to better device performance, with other things equal.Mobility:

5. Temperature coefficient of resistivity (): It is defined as the increase in resistivity per unit resistivity per degree rise in temperature. It is given by = . SI is 0C-1 and dimensions [K-1].Thus, T = 0 [1+ (T-T0)], where T is the resistivity at temperature T and 0 is resistivity at reference temperature T0. Also, RT = R0 [1+ (T-T0)], where RT and R0 are the resistances at temperature T and T0. These formulas are valid only for metallic conductors.

6. Resistivity of different materials and their temperature dependence:a. Metals (Conductors): i. We know, =m/ne2. In a metal, n is not dependent on temperature and thus the decrease in the value of with rise in temperature causes to increase as we have observed.ii. With increase in temperature, thermal motion of ions increase, i.e., atoms vibrate more vigorously and hence resistivity of a metal increases. iii. Resistivity of metals is generally less.The valence electrons of the metal atoms are responsible for conductivity. iv. Value of is high for conductors or metals.b. Semiconductors and Insulators: i. We know that, =m/ne2. thus depends inversely both on the number n of free electrons per unit volume and on the average time between collisions. ii. As we increase temperature, average speed of the electrons, which act as the carriers of current, increases resulting in more frequent collisions. iii. The average time of collisions , thus decreases with temperature. For insulators and semiconductors, n increases prominently with increase in temperature (to be studied in Ch-14). This increase more than compensates any decrease in so that for such materials, decreases with temperature.iv. The resistivity of semiconductors is moderate and that of insulators is high at room temperature. v. Value of is negative for semiconductors and insulators.c. Alloys: i. Some materials like Nichrome (which is an alloy of nickel, iron and chromium) show infinitesimal change in resistivity with increase in temperature which is represented by the beside graph where for a large change in temperature, the change in resistivity is very small.ii. The graph (B) represents very small change in temperature, thus the graph is constant, since the change in resistivity is negligible.iii. Manganin and constantan have similar properties. These materials are thus widely used in wire bound standard resistors since their resistivity values would change very little with temperatures. iv. is very small for alloys of the order of 10-5, thus its slope is also very small. graph (B)S

7. Consider three resistors R1, R2 and R3 connected in series.When a potential difference V is applied across the combination,the same current I flows through each resistance. By Ohms law, the potential drops across the resistance isV1=IR1, V2=IR2, V3=IR3 If RS is the equivalent resistance of the series combination then, V = IRS.But V is equal to the sum of the potential drops across the individual resistances.V = V1 + V2 +V3 Or IRS = IR1+IR2+IR3 [From eq ] Or RS = R1+R2+R3The equivalent resistance of n resistors will be,RS = R1+R2++RnResistors in Series:

8. Properties of Series circuit:i. Current through each resistor is constant.ii. Potential drop across each resistor, Vi Ri and sum of potential drop across each resistor=Total potential difference of battery.iii. All electric components are operated together.iv. For n equal resistors of resistance R connected in series, RS = nR. 9. Resistors in Parallel:

Consider three resistance R1, R2 and R3 connected in parallel between pts A and B. Let V be the potential difference applied across the combination.Let I1, I2 and I3 be the currents through the resistances R1, R2 and R3. Then the current in the main circuit must be:I = I1 + I2 + I3Since all the resistance are connected between same two pts, the potential drop V issame across each. Now by Ohms law:I1= , I2= , I3= If RP is the equivalent resistance, I = , But, I = I1 + I2 +I3 Or = + + [From eqn &] Or = + + For n number of resistances, = + + +

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10. Properties of Series circuit:i. Current through each resistor is constant.ii. Current across each resistor, Ii 1/Ri and sum of currents across each resistor=Total current of battery.iii. For n equal resistors of resistance R each connected in parallel, RP = R/n.

11. Important Points for Combination of Resistors:i. When n resistors of different resistances are connected in series, then the equivalent resistance is maximum and it is greater than any largest individual resistance.ii. When n resistors of different resistances are connected in parallel, then the equivalent resistance is minimum and it is less than the smallest individual resistance.iii. When resistors are connected in series, then their net length is visualised to increase and since R l, hence resistance ses.iv. For resistors connected in parallel, effective area of cross-section is visualised to increase, and since R1/A, hence resistance ses.12. Types of Resistor Combination:1. In the given configuration, R2 and R3 are in parallel (current divides at point B or resistor ends of R2 and R3 meet at common point), and their combination is in series with R1.Thus, equivalent resistance of R2 and R3 is = =>R23= .Now, R23 and R1 are in series. Hence, equivalent resistance of the circuit, Req = R1 + .2. In the given problem, 6 and 3 are in parallel and their combination is in series with 4. Following steps, show the voltage drop and current passing through each resistor.

Homework Sheet 2Total Marks: 34Level 1 (13 marks)Theory (8 marks)1. Write a relation between current and drift velocity of electrons in a conductor. [Compartment 2013, 1mark ]2. Define mobility. Write its dimensions. Write its relationship with relaxation time. [Foreign 2010, Udgam,2 marks]3. Differentiate conductors, semiconductors and insulators on the basis of their resistivity with example. [MAV,2 marks]4. What happens to resistivity of conductor and semiconductors on increasing temperature? Explain with reason for each of them?[2 marks]5. Define the term temperature co-efficient of resistivity. Write its SI units.[1 mark]Graph Based (2marks)6. Draw the graphs showing the variation of resistivity with temperature for i) nichrome ii) silicon. [KV 2012, 2marks]Application Based (3 marks)7. Show that resistance decrease in case of series combination whereas decrease in case of parallel combination? [2 marks]8. *The electron drift speed is established to be only a few mm s-1 for currents in the range of a few amperes. How then is current established almost the instant a circuit is closed?[1 mark]Level 2 (21 marks)Derivations (3 marks)1. Derive the expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time. [Delhi 2013, KV 2012, 3marks]Application Based (3 marks)2. *What is the nature of the path of electrons between two successive collisions with positive ions of metal in the (i) absence of electric field, (ii) presence of electric field?[KV 2012, 1 mark]3. *When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of metal are moving in same direction? Explain.[Udgam, KV, 1mark] 4. *If the electron drift speed is so small, and electrons charge too is small, then how can we obtain large amounts of current in a conductor?[1mark]Numerical (15 marks)5. A current of 5A is passing through a metallic wire of cross section area 410-6m2. If the density of the charge carrier in the wire is 51026m-3, find the drift speed of the electrons.[2marks]Ans. 1/64 ms-1.6. A uniform copper wire of length 1m and cross sectional area 510-7m2 carries a current of 1A. Assuming that there are 81028 free electrons per m3 in copper, how long will an electron take to drift form one end to the other?[Udgam, 2marks]Ans. 6.4103s.7. An aluminium wire of diameter 2.5 mm is connected in series with a copper wire of diameter 1.6mm. A current 2A is passed through them. Find (a) current density in aluminium wire, (b) drift velocity of electrons of copper wire. Number density in copper wire is 1029m-3.[2 marks]Ans. 40.8 MAm-2, 6.210-5 ms-1.8. A 10C of charge flows through a wire in 5 minutes. The radius of the wire is 1mm. It contains 51022 electron/cm3. Calculate the current and drift velocity.[Prakash,2marks] Ans. 3.3310-2A, 1.32610-6 ms-1.9. *A uniform wire is cut into four segments. Each segment is twice as long as the earlier segment. If the shortest segment has a resistance of 4, find the resistance of the original wire.[2 marks]Ans. 60.10. *A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds at a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27 C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.710-4 /0C. [Delhi 2013, 3marks]Ans.762.29 0C. 11. At what temperature would the resistance of a copper conductor be double of its value at 00C? Temperature coefficient of resistance of copper is 410-3/0C. [2 marks]Ans. 2500 C.Session 31. Electromotive Force (emf): It is defined as maximum potential difference between terminals of a battery when no current is drawn from the cell. Its SI unit is Volt (V).2. Electrochemical cell: It is an instrument used to convert chemical energy to electrical energy. It consists of two electrodes, anode (P in the diag): +ve charged and cathode (N in the diag): -ve charged. The potential difference between these two electrodes is stored as the emf of the cell.The potential difference between two electrodes P and N is V+ - (-V-) = V+ + V- =E, emf of cell. 3. Internal Resistance: The opposition offered by electrolyte of a cell to the flow of charge carriers of the electrolytes is called internal resistance of the cell. It depends on the following factors: (i) Increases with decrease in temperature, (ii) Directly proportional to distance between electrodes, (iii) Inversely proportional to area of electrodes, (iv) Directly proportional to concentration of electrolyte, (v) Nature of electrolyte.4. Terminal Potential Difference: The potential drop across the external circuit when a current is being drawn from the cell is called terminal potential difference (V).

Consider a cell of emf E and internal resistance r connected to an external resistance R. Suppose a constant current I flows through this circuit.From Ohms law, Veff = IReffFor a cell, Veff (net potential difference) = E and Reff (net resistance) = R+r. Thus, E = IR + Ir = I (R+r), Hence, current in circuit is, I = . Therefore, potential difference across R is, V = IR = . Also, V = E - V= E Ir. Thus, from above equation, r = = = R =R.Relation between internal resistance and emf:

i. For an open circuit, R = , hence I = 0. In this case, V = E. Graph for V v/s R is given beside. ii. For Imax , R=0.iii. When battery is being charged, then V = E + Ir, whereas for discharging it is V = E Ir.

5. Differences between EMF and potential difference:

Cells in Series:

Suppose two cells of emfs E1 and E2 and internal resistance r1 and r2 are connected in series between pts A and C. Let I be the current flowing through theseries combination. Let VA, VB and VC be the potentials at points A, B and C. The potential difference across terminals of the two cells is,VAB = VA VB = E1 Ir1 And VBC = VB VC = E2 Ir2 Thus the potential difference between pts A and C isVAC = VA VC = (VA VB) + (VB VC) = (E1 Ir1) + (E2 Ir2) [From & ) Or VAC = (E1 + E2) I(r1-r2)VAC = Eeq Ireq, where Eeq=E1+E2 and req=r1+r2In case of n number of cells, Eeq = E1 + E2 + .. + En & req = r1+r2+..+rnIn the above case, if E2 is reversed, then Eeq = E1 E2.

Note: (i) The emf of the battery is the sum of the individual emfs. (ii) The current in each cell is the same and is identical with the current in the entire arrangement. (iii) The total internal resistance of the battery is the sum of the individual internal resistances.(iv) For n equal cells in series, I = .

6. Cells in Parallel:

Suppose two cells of emfs E1 and E2 and internal resistance r1 and r2are connected in parallel between pts A and C. Let currents I1 and I2 fromthe positive terminals of the two cells flow towards the junction B1 and B2 and current I flows out. I = I1 + I2Since both cells between same two points, the potential difference across both cells is same. V between terminals of E1 is V = - =E1-Ir1 I1 = V between the terminals of E2 is V = - = E2-Ir2 I2 = Hence, I = I1+I2 = + [From & ] = (+) V (+) Or V() = () I V = () - I() V = Eeq Ireq, where Eeq = and req = For a parallel combination of n cells, Eeq = + + + and req = + + + In case, E2 battery is reversed, then Eeq =

NOTE: (i) The reciprocal of the total internal resistance is the sum of the reciprocals of the individual internal resistances.(ii) For n equivalent cells is parallel, I = 7. Kirchhoffs Laws: To understand Kirchhoffs laws we must first understand few other terms:i. Junction: Any point in an electric circuit where two or more conductors are joined together.ii. Loop: Any closed conducting path in a circuit.iii. Branch: Any part of the circuit that lies between two junctions.Kirchhoffs first law or Current law or Junction rule: In an electric circuit, the algebraic sum of currents at any junction is zero OR the sum of currents entering a junction is equal to the sum of currents leaving that junction. Mathematically it is expressed as =0.Sign Convention: (i) Current flowing towards junction is taken as positive. (ii) Current flowing away from junction is taken as negative.This law is based on conservation of charges.Kirchhoffs second law or Voltage law or Loop rule: For a closed loop of network, the algebraic sum of changes in potential must be zero OR the algebraic sum of emfs in any loop of a circuit is equal to the sum of the products of currents and resistances in it. Mathematically, it is expressed as =0 or =.

Homework Sheet 3Total Marks: 38Level 1 (12 marks)Theory (7 marks)1. Define EMF and internal resistance of cell.[2 marks]2. Write differences between electromotive force and terminal voltage of a cell?[2 marks]3. State two Kirchhoffs rules.[2 marks]4. State the basic concepts on which two Kirchhoffs laws are based? [KVV,1 marks] Graph Based (2 marks)5. A cell of EMF E and internal resistance r is connected across a variable resistor R. Plot a graph showing the variation of terminal potential V with resistance R. Predict from the graph the condition under which V becomes equal to E.[KVS 2011, MAV, 2marks]Application Based (3 marks)6. Why is the terminal voltage of a cell less than its emf?[1 mark]7. Can Kirchhoffs laws be applied to both d.c. and a.c. circuits?[1 mark]8. Can the terminal potential difference of a cell exceed its EMF? Explain.[1 mark]Level 2 (26 marks)Derivations (7 marks)1. Two cells of EMF E1and E2 and internal resistance r1 and r2 are connected in series. Deduce the expressions for the equivalent EMF and equivalent resistance and current through the circuit.[2 marks]2. A resistance R is connected across a cell of emf E and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell as V. Derive the expression for r in terms of E, V and R. [Delhi 2011, 2 marks]3. Two cells of emfs E1 and E2 and internal resistance r1 and r2 respectively are connected in parallel. Deduce the expressions for (i)equivalent emf, (ii) equivalent resistance of the combination, (iii) Potential difference between the end terminals. [Outside Delhi 2012, Udgam, 3 marks]Application Based (7 marks)4. When cells are connected in parallel, what is the effect on (i) current capacity, (ii) EMF of cells? (increase/decrease). [2 marks]5. When cells are connected in parallel, what is the effect on (i) current capacity, (ii) EMF of cells? [2 marks]6. *A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations: (i) without any external resistance in the circuit, (ii) with resistance R1 only, (iii) with R1 and R2 in series combination, (iv) with R1 and R2 in parallel combination. The currents measured in the four cases are 0.42A, 1.05A, 1.4A and 4.2 A, but not necessarily in that order. Identify the currents corresponding to the four cases mentioned above. [Compartment 2012, Outside Delhi 2012, 3marks]

Numerical (12 marks)7. A network of resistors is connected to a 16 V battery with internal resistance of 1, as shown in Fig.: (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor. (c) Obtain the voltage drops VAB, VBC and VCD.[3 marks] Ans. 7; 1A, 1A,2A, 2/3A, 4/3A; 4V, 2V, 8V8. A battery of EMF 6.6V can supply a current of 3A through a resistance of 1.8, what current does it supply through another resistance of 2.9. Also calculate internal resistance of the battery. [2 marks]Ans. 2A, 0.4.9. A battery of EMF E and internal resistance r gives a current of 0.4A with an external resistor of 12 and a current of 0.25A with an external resistor of 20. Calculate (i) internal resistance, (ii) EMF of battery. [2 marks] Ans. 4/3; 16/3V10. *Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V. When the terminals of the cell are also connected to a resistance of 5, the voltmeter reading drops to 1.8V. Find the internal resistance of the cell. [Outside Delhi 2010, 3 marks] Ans. 10/9.11. *A wire of resistance 8R is bent in the form of a circle. What is the effective resistance between the ends of the diameter AB? [Delhi 2010, DPS, 2 marks]Ans. 2RSession 41. Wheatstone Bridge: Wheatstone Bridge (Direct Current Bridge) is an instrument that is used to measure the resistance or change in resistance and converts it to output current Construction: It consists of four resistances P, Q, R and S; connected to form the arms of a quadrilateral ABCD. A battery of emf E is connected between points A and C and a sensitive galvanometer between B and D. Balanced State: Let S be the resistance to be measured. The resistance of all the resistors is so adjusted that there is no deflection in the galvanometer. The bridge is said to be balanced when the potential difference across the galvanometer is zero so that there is no current through the galvanometer. In the balanced condition of the bridge, = .

Consider a Wheatstone bridge consisting of four resistors P, Q, R and S. Across the pair of diagonally opposite pts, source is connected, called the battery arm. BD is the galvanometer arm. We assume that the cell has no internal resistance. Onapplying Kirchhoffs first law, the currents through various junctions are as shown in figure.Applying Kirchhoffs junction rule to loop ABDA, we get I1P + IgG I2R = 0, where G is the resistance of the galvanometer,Again applying junction rule to loop BCDB, we get(I1-Ig)Q (I2 + Ig)S IgG = 0.Derivation of balanced condition from Kirchhoffs laws:

In the balanced condition of the bridge, Ig = 0. Thus, the above equations become, I1P = I2R I1Q = I2S On dividing eqn by , we get = This proves the condition for balanced Wheatstone bridge.

2. Sensitivity of Whetstone Bridge: A Wheatstone bridge is said to be sensitive if it shows a large deflection in the galvanometer for a small change of resistance in the resistance arm.i. Wheatstone bridge is most sensitive when all the resistance of the four arms are of same order. It is mainly used to find resistances of medium value because for low resistances end-resistances, resistances of connecting wires become comparable to resistance being measured and hence introduce error in the result.Whereas for measuring high resistance, all other resistance should also be high to ensure sensitivity. But this reduces the current through the galvanometer which then becomes insensitive.ii. In Wheatstone bridge, if we exchange the battery and the galvanometer, then it does not have any effect on the sensitivity of the bridge since condition for balance of the bridge remains satisfied. Thick copper wires should be used for connections in a Wheatstone bridge since resistance 1/A, hence, resistance of wires will be low so that it does not cause any error in the calculation of unknown resistance.Advantage of using Wheatstone Bridge for measuring resistance is that since it is a null method, hence it doesnt considers the cell resistances of the circuit. Cases of Wheatstone Bridge: In the given arrangement, the system is like a wheatstone bridge where R1 and R5 form P and Q whereas R3 and R4 form R and S and there is no current through R2.

3. Meter Bridge:

It is the simplest practical application of Wheatstone bridge that is used to measure an unknown resistance. Principle: Its working is based on the principle of Wheatstone bridge.When the bridge is balanced, = .Construction: It consists of usually one metre long manganin wire of uniform cross section, stretched along a metre scale fixed over a wooden board and with its two ends soldered to two L-shaped thick copper strips A and C.A resistance R is connected in the gap ab and the unknown resistance S is connected in the gap a1b1. A source of emf E is connected across AC. A movable jockey and a galvanometer are connected across BD.

Working: After taking out a suitable resistance R from the resistance box, the jockey is moved along the wire AC till there is no deflection in the galvanometer. This is the balanced condition of the Wheatstone bridge.If P and Q are resistances of parts AB and AC, then for balanced condition, = .Let AJ=l cm, then BJ=(100-l) cm. Since the bridge wire is of uniform cross-section, thereforeResistance of wire length of wire. = = = , where is resistance per unit length of wire. Hence, = S = . Knowing l and R, unknown resistance can be determined.

Meter Bridge is most sensitive when the balanced condition or the null point is obtained at the centre of the wire, i.e., at 50cm. This happens when the resistances of R and S are comparable. At the middle of the slide wire or meter bridge, the value of the unknown resistance is most accurate.Homework Sheet 4Total Marks: 24Level 1 (11 marks)Theory (5 marks)1. State the working principle of Wheatstone bridge. [1 mark]2. With a circuit diagram, briefly explain how a metre bridge can be used to find the unknown resistance of a given wire. [KVV 2011, 3 marks]3. When is Wheatstone bridge most sensitive?[1 mark]Application Based (6 marks)4. Can Meter Bridge be used for finding the resistance of (i) moderate values, (ii) high values, (iii) low values? Explain.[2 marks]5. At what positions of the jockey on slide Meter Bridge, the results are most accurate?[1 mark]6. Why the connections between resistors in a Wheatstone bridge are made of thick copper wire? [1 mark]7. In the meter bridge experiment, balance point was observed at J with AJ=l. (i) The values of R and X were doubled and then interchanged. What would be the new position of balance point? (ii) If the galvanometer and battery are interchanged at the balance point, how will the balance point get affected? [All India 2011, KV 2012, KV 2014, Delhi 2013 2 marks]Level 2 (13 marks) Derivations (3 marks)1. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge. [Delhi 2013, MAV, DPS, 3 marks]Numerical (10 marks)2. Determine the current in each branch of the network shown. [3 marks] Ans.4/17A, 6/17A, -4/17A, 6/17A, -2/17A.3. In a meter bridge, the balancing point is found to be at 39.5 cm from resistor Y is of 12.5. Determine the resistance of X. [Delhi 2013, 2 marks]Ans. 8.16.4. With a certain resistance in the left gap of a meter bridge, the balancing point is obtained when a resistance of 10 is taken out from the resistance box. On increasing the resistance from the resistance box by 12.5 , the balancing point shifts by 20cm. Find the value of the unknown resistance.[3 marks]Ans. 15 .5. In comparing the resistance of two coils P and Q with a slide wire bridge, a balance point is obtained when the sliding contact is 30cm from the zero end of the wire. The resistances P and Q are interchanged and the balance is obtained at 120cm from the same end. Find the ratio of the resistances P and Q and the length of the bridge wire.[2 marks]Ans. 1:4, 150cm. Numerical Assignment 1 (26 marks)1. (a) Three resistors 1 , 2 , and 3 are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.[2 marks]Ans. 6, 2V, 4V, 6V.2. Calculate the steady current through the 2 resistor in the circuit shown. [Foreign 2010, 3 marks] Ans. 9/10A.3. Two cells of emf 1.5 V and 2 V and internal resistance 1 ohm and 2 ohm respectively are connected in parallel to pass a current in the same direction through an external resistance of 5 ohm. (a) Draw the circuit diagram. (b) Using Kirchhoffs laws, calculate the current through each branch of the circuit and potential difference across the 5 ohm resistor. [Outside Delhi 2013, 3 marks]Ans. 0.423A, 0.038A, 0.423A (approx.) 4. Two cells E1 and E2 5V and 9V and internal resistances of 0.3 and 1.2 respectively are connected to a network of resistances as shown in the figure. Calculate the value of current flowing through the 3 resistance. [Foreign 2010, 2 marks] Ans. 1/3A5. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2A. What would be the potential difference between the points B and E? [Outside Delhi 2012, KV 2014, 3 marks] Ans. 5. 4. A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 39 as shown in the figure. Find the value of the current. [Delhi 2013, 2 marks] Ans. 5A. 5. Two identical cells, of emf 1.5 V each are joined in parallel providing supply to an external circuit consisting of two resistors of 13 each joined in parallel. A very high resistance voltmeter reads the terminal voltage of the cells to be 1.4 V. Find the internal resistance of each cell. [Foreign 2013, 2 marks] Ans. (13/14)6. *Determine the current drawn from a 12V supply with internal resistance 0.5 by the infinite network. Each resistor has a resistance of 1 each. Ans. 3.7A. [3 marks]7. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 (Fig). Determine the equivalent resistance of the network and the current along each edge of the cube. [3 marks] Ans. R=5/6, Value of I=4A (in fig)

8. Determine the current in each branch of the network shown in Fig. [3 marks] Ans. I1=2.5A, I2=5/8 A, I3=15/8 A

Session 51. It is a device used to measure an unknown emf or potential difference accurately. It is a device which does not draw any current from the battery whose emf is to be measured. Potentiometer is a null method device. At null point, it does not draw any current from the cell and thus there is no potential drop due to internal resistance of the cell. It measures the potential difference in an open circuit which is equal to actual emf of the cell. Because of this potentiometer is also called ideal voltmeter and hence is preferred over voltmeter.Principle: When a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.We know by Ohms Law, V = IR = I..For a wire of uniform cross-section and uniform composition, and A are constants.Therefore, when a steady current I flows through the wire, then = =constant.Hence, V= l => V l.This is the principle of potentiometer.Potential Gradient: The potential drop per unit length of the potentiometer wire is known as potential gradient. It is denoted by or k. It is given by = . SI unit of potential gradient =Vm-1. Practical unit =Vcm-1.Construction: It consists of a long wire AB of uniform cross-section usually 4 to 10m long, of material having high resistivity and low temperature co-efficient such as constantan or manganin. The ends A and B are connected to a strong battery E, a plug key K and a rheostat Rh. This circuit is called driving or auxillary circuit which sends a constant current I through the wire AB. Thus, the potential gradually falls from A to B. A jockey can slide along the length of the wire.Potentiometer:

2. Consider in first case, position of the key where 1 and 3 are connected so that the galvanometer is connected to E1. The jockey is connected on the wire tillthe galvanometer shows no deflection. Let the pt be N1 at a distance l1 from A.In case of loop AN1G31A, applying junction rule,l1+0 E1=0 Similarly, another emf E2 is balanced against l2 (AN2) by closing key between 2 and 3: l2+0 E2=0 From eqn & , = Note: One of the cells chosen is a standard one (whose emf is known) so that the emf of the other cell can be easily calculated.Comparison of emfs of two primary cells:

3. Internal Resistance of the cell:

Consider that a cell with emf E whose internal resistance is r is to be Determined is connected across a resistance box through a key K2. With key open, balance is obtained at l1 (AN1). E = l1When K2 is closed, the cell sends a current (I) through the resistance box. If V is the terminal potential difference of the cell & balance is obtained at l2 V = l2 we have = But, E = I (r+R) and V = IR, = From and , r = R NOTE: A potentiometer draws no current from the voltage source being measure.

4. Sensitivity of potentiometer: A potentiometer is sensitive if it can measure very small potential differences, i.e., it shows a significant change in balancing length for a small change in potential difference.Sensitivity of a potentiometer depends on the potential gradient along its wire. Smaller the potential gradient, greater is the sensitivity of the potentiometer.Sensitivity of a potentiometer can be increased by (i) Increasing the length of the potentiometer wire, (ii) For a fixed length of wire, decreasing the current in the circuit with help of a rheostat.5. Remember Colour Coding: B B R O Y of Great Britain Very Good Wife Gold-5% Silver-10%Black Brown Red Orange Yellow Green Blue Violet Gray White No fourth band = 20%0 1 2 3 4 5 6 7 8 9The first two bands from the end indicate the first two significant figures of the resistance in ohms. The third band indicates the decimal multiplier. The last band stands for tolerance or possible variation in percentage about the indicated values. Sometimes, this last band is absent and that indicates a tolerance of 20%. For e.g. if there are four colour bands- red, orange, blue and gold. Then the resistance of resistor will be 23106 5%.6. Non-Ohmic Conductors: Although Ohms law has been found valid over a large class of materials, but there are materials where V I is not valid. Such types of material are:

7. Ohmic Loss:

Consider a conductor with end points A and B in which a current I is flowing from A to B. The electric potentials at A and B are VA and VB. Since current is flowing from A to B, VA > VB and potential difference across AB isV = VA VB > 0In time t, Q=It travels from A to B. The P.E. at A and B is QVA and QVB. Upot. = Final P.E. Initial P.E. = Q [VB VA] = -QV = -IVt < 0

If charge moved without collision through the conductor, K.E. would change so that the total energy remains unchanged. K = -Upot. i.e., K = IVt >0Thus, K.E. increases as charges move freely.So though K.E. increases, but due to collisions, this energy is lost in the form of heat and thus electrons travel with average drift velocity. In an actual conductor, the amount of energy dissipated as heat during time interval t is W = IVtWe know, P = IV Using Ohms Law, V = IR, we get P = I2R = Thus, the above expression represents power loss (ohmic loss) in a conductor of resistance R carrying a current I.

8. Reason for supplying power at high voltage:

Consider a device R, to which a power P is to be delivered via transmission cables having a resistance RC to be dissipated by it finally.If V is the voltage across R and I the current through it, then P = VIThe connecting wires from the power station to the device has a finite resistance RC. The power dissipated in the connecting wires, which is wasted is PC with PC = I2RC = Thus to a device of power P, the power wasted in the connecting wires is inversely proportional to V.

9.

Homework Sheet 5Total Marks: 37Level 1 (13 marks)Theory (10 marks)1. What do you mean by the sensitivity of potentiometer? How it is related to the potential gradient? [KV 2012, 2 marks]2. On what factors does the potential gradient of potentiometer wire depend?[1 mark]3. What is the reason for power loss in a conductor?[2 marks]4. Why power is transmitted at very high voltages? Explain.[MAV, 2 marks]5. Draw VI graph for ohmic and non-ohmic materials. Give one example for each. [All India 2013, 3 marks]Application Based (3 marks)6. Why do we prefer a potentiometer to measure EMF of a cell rather than a voltmeter?[1 mark]7. Why do we prefer potentiometer of longer length for accurate measurements?[1 mark]8. If the length of the wire be (i) doubled (ii) halved, what will be the effect on the position of zero deflection in a potentiometer? Explain.[1 mark]Level 2 (24 marks)Derivations (11 marks)1. (a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primary cells. Obtain the required expression used forcomparing the emfs. (b)Write two possible causes for one sided deflection in a potentiometer experiment.[5 marks]2. With the help of circuit diagram, describe a method to find the internal resistance of a primary cell. [Delhi 2013, Delhi 2010, MAV, Prakash, 3 marks]3. Two heating elements of resistance R1 and R2 when operated at a constant supply of voltage V consume powers P1 and P2 respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in (i) series, (ii) parallel across the same voltage supply. [Outside Delhi 2011, KVS, 3 marks]Application Based (1 mark)4. If the EMF of the driving cell is decreased, what will be the effect on the position of zero deflection in potentiometer? Explain.[1 mark]Numerical (6 marks)5. A potentiometer having a wire 10m long stretched on it is connected to a battery having steady voltage. A leclanche cell gives a null point at 750cm. If the length of the potentiometer wire be increased by 100cm find the new position of the null point.[2 marks]Ans. 825cm. 6. A voltage of 30 V is applied across a carbon resistor with first, second and third rings of blue, black and yellow colours respectively. Find the value of current through the resistor. [Delhi 2013, 1 mark]Ans. 50A.7. (a) You are required to select a carbon resistor of resistance 47k10% from a large collection. What should be the sequence of colour bands used to code it? (b) Write characteristics of manganin which make it suitable for making standard resistance. [Foreign 2011, Udgam, 2 marks]8. A current of 2mA is passed through a colour coded carbon resistor with first, second and third rings of yellow, green and orange colours. What is the voltage drop across the resistors? [1 mark]Ans. 90V.Value Based (6 marks)9. Rakesh purchased cells for his transistor. He felt that cells are not working properly. He wanted to check their e.m.f. So, he took the cell to the physics lab and with the help of potentiometer found their e.m.f. To his surprise e.m.f. was less than the value claimed by the manufacturer. He lodged the complaint with consumer forum and received the deserving response. a. What values are displayed by Rakesh? b. What do you think why Rakesh used potentiometer instead of voltmeter to find out e.m.f. of the cell? For more precise measurement the potential gradient of the potentiometer should be high or low?[3 marks]10. Laxmi and her mother went to the market to purchase some household articles. Laxmis mother was going to purchase 100 W electric bulb. Laxmi advised her to purchase CFL. She told her mother that it will consume less amount of power and will save electricity. i. What qualities do you notice in Laxmi? ii. A 100 W bulb and a 500 W bulb are joined in parallel to the mains. Which bulb will draw more current?[3 marks]Numerical Assignment 2Total Marks: 381. A heating element is marked 210 V, 630 W. Find the resistance of the element when connected to a 210 V dc source. [Delhi 2013, 2 marks] Ans. 70.2. In a potentiometer arrangement, a cell of EMF 1.2V gives a balance point at 30cm length of the wire. This cell is now replaced by another cell of unknown EMF. If the ratio of EMFs of the two cells is 1.5, calculate the difference in the balancing length of the potentiometer wire in the two cases.[2 marks]Ans. 10cm.3. A standard cell of EMF 1.08V is balanced by the potential difference across 91cm of a metre long wire supplied by a cell of EMF 2V through a series resistor of resistance 2. The internal resistance of the wire is zero. Find the resistance per unit length of the wire.[3 marks]Ans. 0.03cm-1.4. *A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds at a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27 C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.710-4 C0. [Delhi 2013, 3 marks]Ans.762.29 0C. 5. The resistance of a conductor is 6 at 50 C and 7 at 100 C. Calculate the temperature coefficient of resistance of the material. Find the resistance of the conductor at 0 0C. [2 marks] Ans. 3.3310-3C-1, 5.143.6. A parallel combination of three resistors takes current of 7.5A from a 30V supply. If the two resistors are 10 and 12, find the third one. [2 marks]Ans. 157. *We have n resistors each of resistance r. These are first connected to get minimum resistance and then again connected to get maximum resistance. Compute the ratio of resistances in the two cases. [1 mark]Ans.1:n2.8. Two wires X, Y have the same resistivity, but their cross-sectional areas are in the ratio 2:3 and lengths in the ratio 1:2. They are first connected in series and then in parallel. Find out the ratio of drift speeds of electrons in the two wires for two cases. [KV 2011, 2 marks]Ans. 3:2, 2:1.9. A resistor of 5 is connected in series with a parallel combination of a number of resistors each of 5. If the total resistance of the combination is 6. How many resistors are in parallel?[2 marks]Ans. 5.10. A 10m long wire of uniform cross section and resistance 20 resistance is used in a potentiometer. The wire is connected in series with a battery of 5V along with an external resistance of 480. If an unknown EMF E is balanced at 6m of the wire, calculate (i) potential gradient of the wire, (ii) the value of unknown EMF E. [Delhi 2006, 3 marks]Ans. 0.02Vm-1, 0.12V.11. In a potentiometer circuit for comparison of two resistances, the balance point with a standard resistance R=10 is at 58.3cm while with that of unknown resistance X is 68.5cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of EMF E?[2 marks]Ans. 11.75. 12. In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emf E1 and E2 connected in the manner shown are obtained at a distance of 120cm and 300 cm from the end A. Find (i) E1/E2 and (ii) position of null point for the cell E1. How is the sensitivity of the potentiometer increased? [Outside Delhi 2012, 3 marks] Ans. 7:3, 210cm 13. Using Kirchoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 resistance. Also find the potential difference between A and D. [Compartment 2012, Outside Delhi 2012, 3 marks] Ans. 2, 3V.14. The supply voltage to a room is 120V. The resistance of the lead wires is 6. A 60W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240W heater is switched on in parallel to the bulb? (a) 10.04 Volt, (b) 0 Volt, (c) 2.9 Volt, (d) 13.3 Volt. [JEE Main 2013, 4 marks] Ans. (a)15. Two electric bulbs marked 25W 220V and 100W 220V are connected in series to a 440Vsupply. Which of the bulbs will fuse? (a) both (b) 100 W (c) 25 W (d) neither. [AIEEE 2012][4 marks]Ans. (c)

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