In conclusion, there are two requirements which must be met in order to establish an electric circuit. The requirements are:

1.There must be an energy supply capable doing work on charge to move it from a low energy location to a high energy location and thus creating an electric potential difference across the two ends of the external circuit. 2.There must be a closed conducting loop in the external circuit which stretches from the high potential, positive terminal to the low potential negative terminal.

RESISTORS IN SERIES

- In a series circuit, the current is the same at all points along the wire.  IT = I1 = I2 = I3

 - An equivalent resistance is the resistance of a single resistor that could replace all the resistors in a circuit. The single resistor would have the same current through it as the resistors it replaced.  RE = R1 + R2 + R3

 - In a series circuit, the sum of the voltage drops equal the voltage drop across the entire circuit.

VT = V1 + V2 + V3

RESISTORS IN SERIES

12.4 Two resistances of 2 Ω and 4 Ω respectively are connected in series. If the source of emf maintains a constant potential difference of 12 V. a. Draw a schematic diagram with an ammeter and a voltmeter.

12.4 Two resistances of 2 Ω and 4 Ω respectively are connected in series. If the source of emf maintains a constant potential difference of 12 V. b. What is the current delivered to the external circuit?

Re = R1 + R2

= 2 + 4 = 6 Ω

IV

RTe

12

6= 2 A

c. What is the potential drop across each resistor?

V1 = I R1

= 2(2) = 4 VV2 = I R2

= 2(4) = 8 V

PARALLEL CIRCUITS

- In a parallel circuit, each resistor provides a new path for electrons to flow. The total current is the sum of the currents through each resistor.  IT = I1 + I2 + I3

- The equivalent resistance of a parallel circuit decreases as each new resistor is added.  

- The voltage drop across each branch is equal to the voltage of the source.

VT = V1 = V2 = V3

1 1 1 1

1 2 3R R R RE

RESISTORS IN PARALLEL

KIRCHHOFF’S LAWS An electrical network is a complex circuit consisting of current loops. Kirchhoff developed a method to solve this problems using two laws.

Law 1. The sum of the currents entering a junction is equal to the sum of the currents leaving that junction.

I Iin out

This law is a statement of charge conservation.

A junction (j) refers to any point in the circuit where two or three wires come together.

j

KIRCHHOFF’S LAWS

Law 2. The sum of the emfs around any closed current loop is equal to the sum of all the IR drops around that loop. (Ohm’s Law: V = IR)

This law is a statement of energy conservation.

IR

Gustav Robert Kirchhoff             (1824-1887)

12.5 The total applied voltage to the circuit in the figure is 12 V and the resistances R1, R2 and R3 are 4, 3 and 6 Ω respectively.

a. Determine the equivalent resistance of the circuit.

R2 and R3 are in parallel (RP)

1 1

6

1

3Rp Rp= 2 Ω

RP and R1 are in series

Re = 4 + 2

= 6 Ω

b. What is the current through each resistor?

IV

RTe

12

6= 2 A

I1 = 2 A (series)

V1 = I1R1

= 2(4) = 8 V

The voltage across the parallel combination is therefore: 12 - 8 = 4 V each

IV

R22

4

3= 1.33 A I

V

R33

4

6= 0.67 A

12.6 Find the equivalent resistance of the circuit shown.

1 and 2 are in series:1+ 2 = 3 Ω

this combination is in parallel with 6:

this combination is in series with 3: 2 + 3 = 5 Ω

1 1 1

3 6pR

this combination is in parallel with 4:

1 1 1

5 4pR RP = 2.22 Ω = Req

RP = 2 Ω

12.7 A potential difference of 12 V is applied to the circuit in the figure below. a. Find the current through the entire circuit

R1 and R2 are in parallel:

1 1 1

4 4pR RP = 2 Ω

This combination is in series with R3:

2 + 4 = 6 Ω

This combination is now in parallel with R4:

= 3 Ω = Req

1 1 1

6 6pR

IV

RTeq

= 4 A

IT = I3 + I4

IV

R44

= 2 A

I3 = IT - I4

= 4 - 2 = 2 A

b. Find the current through each resistor.

12

3

12

6

The voltage for the parallel combination is:V' = V - I3 R3

= 12 - (2)(4) = 4 V

IV

R11

'

= 1 A I2 = 2 - 1 = 1 A4

4

EMF AND TERMINAL POTENTIAL DIFFERENCE Every source of emf (Є) has an inherent resistance called internal resistance represented by the symbol r. This resistance is a small resistance in series with the source of emf. The actual terminal voltage VT across a

source of emf with an internal resistance is given by: 

VT = Є - I r Units: Volts (V)

12.8 A load resistance of 8 Ω is connected to a battery whose internal resistance is 0.2 Ω a. If the emf of the battery is 12 V, what current is delivered to the load?

Є = 12 VRL = 8 Ω

r = 0.2 Ω

IV

R

12

8 0 2.= 1.46 A

b. What is the terminal voltage of the battery?

VT = Є - I r

= 12 - 1.46(0.2) = 11.7 V

LR r

12.9 a. Determine the total current delivered by the source of emf to the circuit in the figure. V = 24 V. The resistances are 6, 3, 1, 2 and 0.4 Ω respectively.

this combination is in series with R3:

2 + 1 = 3 Ω

1 1

6

1

3Rp

R1 and R2 are in parallel:

this combination is now in parallel with R4:

1 1

2

1

3Rp RP = 1.2 Ω

RP = 2 Ω

finally the internal resistance r is in series giving the equivalent resistance:

Req = 1.2 + 0.4

= 1.6 Ω

IV

RTeq

24

16.= 15 A

b. What is the current through each resistor?

VT = Є - I r

= 24 - 15(0.4) = 18 V

V4 = VT = 18 V

IV

R44

18

2= 9 A

I3 = IT - I4

= 15 - 9 = 6 A

V3 = I3R3

= 6(1) = 6 V

V1 = V2 = 18 - 6 = 12 V each

IV

R11

12

6= 2 A I

V

R22

12

3= 4 A

AmmeterAmmeterVoltmeterVoltmeter RheostatRheostatSource Source of EMFof EMF

Rheostat

A

Ammeters and Voltmeters

V Emf-

+

Galvanometer

00 10102020

10102020

N S

The galvanometer uses torque created by small currents as a means to indicate electric current.

A current Ig causes the needle to deflect left or right. Its resistance is Rg.

The sensitivity is determined by the current required for deflection. .

Operation of an Ammeter

The galvanometer is often the working element of both ammeters and voltmeters.

A shunt resistance in parallel with the galvanometer allows most of the current I to bypass the meter. The whole device must be connected in series with the main circuit.

I = Is + IgI = Is + Ig

RRgg

II RRss

IIss

IIgg

The current Ig is negligible and only enough to operate the galvanometer. [ Is >> Ig ]

Operation of an Voltmeter

The voltmeter must be connected in parallel and must have high resistance so as not to disturb the main circuit.

A multiplier resistance Rm is added in series with the galvanometer so that very little current is drawn from the main circuit.

VB = IgRg + IgRmVB = IgRg + IgRm

RRgg

II

VVBB

IIgg

The voltage rule gives:

RRmm

CAPACITORS IN SERIES AND PARALLEL

These are the symbols used in different arrangements of capacitors:

CAPACITORS IN SERIES

...C1

C1

C1

C1

321eq

....QQQQ 321T

....VVVV 321

CAPACITORS IN SERIES

Series capacitors always have the same charge. The voltage across the equivalent capacitor Ceq is the sum

of the voltage across both capacitors.

CAPACITORS IN PARALLEL

....CCCC 321eq

....QQQQ 321T

....VVVV 321

CAPACITORS IN PARALLEL

Parallel Capacitors always have the same voltage drop across each of them. The charge on the equivalent capacitor Ceq is the sum of the

charges on both capacitors.

19.9 a. Find the equivalent capacitance of the circuit.

C2 and C4 are in series

1 1 1

2 4 2 4C C C,

CC C

C C2 42 4

2 4,

2 4

2 4

( )= 1.33 μF

C2 = 2 μF, C3 = 3 μF, C4 = 4 μF

V = 120 V

C3 is now in parallel with C2,4

Ceq= C3 + C2,4

= 3 +1.33 = 4.33 μF

b. Determine the charge on each capacitor.

The total charge of the systemQT = CeqV

= 4.33 (120) = 520 μC

Q3 = C3V

= 3(120) = 360 μC

Q2 and Q4 have the same charge since they are in series:

Q2 = Q4 = QT - Q3

= 520 - 360 = 160 μC

Q3 = 360 μC, Q2 = Q4 =160 μC

c. What is the voltage across the 4 μF capacitor?

VQ

C44

4

160

4= 40 V

The remaining voltage (120 - 40 = 80 V) goes through the C2 capacitor.

RC CIRCUITS A resistance-capacitance (RC) circuit is simply a circuit containing a battery, a resistor, and a capacitor in series with one another. An RC circuit can store charge, and release it at a later time. A couple of rules dealing with capacitors in an RC circuit:

1. An empty capacitor does not resist the flow of current, and thus acts like a wire.

2. A capacitor that is full of charge will not allow current to flow, and thus acts like a broken wire.

When the switch is closed, the capacitor will begin to charge.

RC Circuits

If an isolated charged capacitor is connected across a resistor, it discharges:

RC Circuits

19.10 Three identical resistors, each with resistance R, and a capacitor of 1.0 x 10‑9 F are connected to a 30 V battery with negligible internal resistance, as shown in the circuit diagram above. Switches SI and S2 are initially closed, and switch S3

is initially open. A voltmeter is connected as shown.

a. Determine the reading on the voltmeter.

b. Switches Sl and S2 are now opened, and then switch S3 is

closed. Determine the charge Q on the capacitor after S3 has

been closed for a very long time.

After the capacitor is fully charged, switches S1 and S2 remain

open, switch S3 remains closed, the plates are held fixed, and

a conducting copper block is inserted midway between the plates, as shown below. The plates of the capacitor are separated by a distance of 1.0 mm, and the copper block has a thickness of 0.5 mm.

 c. i. What is the potential difference between the plates?

ii. What is the electric field inside the copper block?

iii. On the diagram, draw arrows to clearly indicate the direction of the electric field between the plates.

iv. Determine the magnitude of the electric field in each of the spaces between the plates and the copper block.