7. D.C. CIRCUITS Content Learning Outcomesjjphysics.pbworks.com/f/10_11_H1_DC_notes_tr.pdf(b) draw and interpret circuit diagrams containing sources, switches, resistors, ammeters,

8866 H1 Physics – J2/2011 7. D.C. Circuits

7. D.C. CIRCUITS Content • Practical circuits • Series and parallel arrangements Learning Outcomes Candidates should be able to: (a) recall and use appropriate circuit symbols as set out in SI Units, Signs,

Symbols and Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995).

(b) draw and interpret circuit diagrams containing sources, switches, resistors,

ammeters, voltmeters, and/or any other type of component referred to in the syllabus.

(c) solve problems using the formula for the combined resistance of two or more

resistors in series. (d) solve problems using the formula for the combined resistance of two or more

resistors in parallel. (e) solve problems involving series and parallel circuits for one source of e.m.f.

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Types of electric current

• Direct Current (D.C.): Flow of charges in the circuit is in the same direction all the time, from a higher potential to a lower potential (e.g. current from battery).

• Alternating Current: Flow of charges in the circuit reverses direction at

regular intervals (e.g. current from household mains). Electric circuits consist of circuit components (e.g. batteries, resistors, and switches) connected by conductors (e.g. copper cables). For electric current to flow, the circuit components and conductors must form closed loops. There must also be sources of electrical energy (e.g. batteries) and sinks of electrical energy among the circuit components (e.g. resistors and lamps). (a) recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols

and Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995).

Electrical Circuit Symbols Symbol modifier

(a) Recall and use

variable value inherent non-linearity preset value Conductors and terminals

with connection between conductors no connection between conductors

current-carrying conductor open terminals ground/earth

single pole single throw (SPST) switch

single pole double throw (SPDT) switch

double pole double throw (DPDT) switch

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D.C. sources

cell battery (>1 cell) photovoltaic cell

d.c. source variable d.c. source Measuring instruments

ammeter and milliammeter voltmeter and millivoltmeter

galvanometer (alphabetical symbol, null deflection, current detected in certain direction, current detected in another direction)

Resistors

fixed resistor variable resistor (rheostat) thermistor

light-dependent resistor (LDR) potentiometer (voltage divider) Other circuit components

diode photodiode light-emitting diode (LED)

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fuse heating element indicator, lamp or light source Other symbols found in TYS

generator motor loudspeaker thermocouple

oscilloscope aerial/antenna electric bell buzzer (b) draw and interpret circuit diagrams containing sources, switches, resistors,

ammeters, voltmeters, and/or any other type of component referred to in the syllabus.

Note that, for a certain electric circuit, there are different ways of drawing its circuit diagram.

(b) Draw and interpret

Actual circuit Circuit diagram Two other possible circuit diagrams for the above electric circuit are as follows:

V

R1

R2

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Example 1: Draw the circuit diagram for the electric circuit shown below.

Actual circuit Circuit diagram Conservation of charge (must know in order to solve circuit problems) Given that we are dealing with steady currents (i.e. no accumulation of charge at circuit junctions), the sum of currents entering a circuit junction is equal to the sum of currents leaving it.

54321 IIIII ++=+ Taking currents entering circuit junction as positive and currents leaving circuit junction as negative, we have:

054321 =−−−+ IIIII Taking currents leaving circuit junction as positive and currents entering circuit junction as negative, we have:

021543 =−−++ IIIII Example 2 The given diagrams show wires carrying currents I1, I2, I3, and I4, meeting at a junction. Which of the following diagrams represents the equation I1 + I2 = I3 + I4?

A B C D

I1 + I3 + I4 + = I2 I1 + I2 + I3 = I4 I1 + I2 = I3 + I4 (correct answer)

Impossible, all currents are entering

and no current leaving.

Conservation of energy (must know in order to solve circuit problems)

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The algebraic sum of e.m.f. (i.e. sources of electrical energy) is equal to the algebraic sum of p.d. (i.e. sinks of energy) for any closed loop within the circuit. Example 3

Find I1 and I2 in terms of E1, E2, R1, R2 and R3. Given that E1 = 3.0 V E2 = 1.5 V R1 = R2 = R3 = 10 Ω find the values of I1 and I2.

Solution: Conservation of energy for the loop:

( )211211121 RRIRIRIEE +=+=+

21

211 RR

EEI++

=

Conservation of energy for the loop:

3221 RIEE =+

3

212 R

EEI +=

Substituting E1 = 3.0 V, E2 = 1.5 V and R1 = R2 = R3 = 10 Ω, I1 = 0.23 A and I2 = 0.45 A.

(c) solve problems using the formula for the combined resistance of two or more

resistors in series. Derivation of the effective resistance of resistors in series (for reference only):

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For series connection, the effective resistance is the sum of individual resistance. In general, for n similar resistors (each of resistance R) connected in series, the effective resistance is . nRReff = Note: The effective resistance always increases when additional resistors are

connected in series.

1. Same water current flowing through each

section of the tube, whether wide or narrow (water cannot be compressed, no accumulation, inflow = outflow).

2. Resistance to water flow increases as number of narrow portions along the tube increases (adding obstruction to water flow).

Example 4 Calculate the effective resistance of a 4 Ω and two 3 Ω resistors connected in series.

Solution:

Ω=+= 10)3)(2(4effR

(d) solve problems using the formula for the combined resistance of two or more

resistors in parallel. Derivation of the effective resistance of resistors in parallel (for reference only):

For parallel connection, the reciprocal of effective resistance is the sum of reciprocal of individual resistance.

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In general, for n similar resistors (each of resistance R) connected in parallel, the

effective resistance is Rn

Reff

=1

nRReff =⇒ .

Note: The effective resistance always decreases when additional resistors are

connected in parallel.

The effective resistance of resistors in parallel is always less than the individual resistance of each resistor.

1. The water current through the wide tube

is the same as the sum of the water currents in each of the narrow tubes (water cannot be compressed, no accumulation, inflow = outflow).

2. Resistance to water flow decreases as number of narrow tubes increases (adding channels for water flow).

Example 5 Calculate the effective resistance of a 2 Ω, a 3 Ω and a 4 Ω resistor connected in parallel. Solution:

41

31

211

++=effR 12

346 ++=

1213

=

Ω= 903.0effR

(e) solve problems involving series and parallel circuits for one source of e.m.f.

Example 6 A battery C of 1.5 V and negligible internal resistance is connected to the combination of resistors as shown. Find the values of the currents I1, I2, and I3 in the diagram.

109.075.310

5.11 =

+=I

11.0= A

23

23

3155

IIII

=⇒

=

222321 43 IIIIII =+=+=

10 Ω 10 Ω

5 Ω 5 Ω

I2

I3 CI1

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10 Ω

0273.04109.0

41

2 ===⇒II

027.0= A

213 III −= 0817.00273.0109.0 =−= 082.0= A

3.75 Ω CI1

More worked examples Find the equivalent resistances between points A and point B in Examples 7, 8, and 9. Assume that all resistances in the circuits are 2 Ω each. Example 7

A

B E

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Solution: Step 1: Assign potential at each terminal of each resistors. Step 2: Group resistors which have same potential at its terminals. Step 3: Simplify the circuit

VA

VF

VB

VD

2 Ω //2 Ω = 1 1

1 12 2

= Ω+

2 Ω

2 Ω 1Ω 1.33 Ω

E

VA

VB E

VA VC

VF VD

VA VB

E VA VC

VF VD

2 //4 Ω Ω1 1.33

1 12 4

= Ω+

=

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VA VB 3 //2 Ω Ω

1 1.2 1 13 2

= =+

Ω 1.33 Ω

Hence, effective resistance between point A and point B = 1.2 + 1.33 = 2.53 Ω Example 8 Solution: Step 1: Assign potential at each terminal of each resistors

VD 1.2 Ω

A

B

E

VC

VD VA

VB VC

VF

E

E

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Step 2: Group resistors which have same potential at its terminals. Step 3: Simplify the circuit

Hence, effective resistance between point A and point B = 2 + 1.2 = 2.2 Ω Example 9

VA

VB

VF VD

E

2 Ω //2 Ω //4 Ω1 0.8

1 1 12 2 4

= = Ω+ +

VC VC

//(2 Ω + 0.8 Ω ) Ω21 1.2

1 12 2.8

= = Ω+

A B E

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Solution: Step 1: Assign potential at each terminal of each resistors Step 2: Group resistors which have same potential at its terminals. Step 3: Simplify the circuit

Hence, effective resistance between point A and point B = 2 Ω

--END--

0 Ω (short circuit)

VB VA

VA VB E

VB VC

VB

VB

VA VB E

VC

VB

VB

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7. D.C. CIRCUITS Content Learning Outcomesjjphysics.pbworks.com/f/10_11_H1_DC_notes_tr.pdf(b) draw and interpret circuit diagrams containing sources, switches, resistors, ammeters,