1 ©

John Parkinson

KIRCHHOFF’S LAWSKIRCHHOFF’S LAWS&&

COMBINATIONS OF RESISTORSCOMBINATIONS OF RESISTORS

•RESISTORS IN SERIES

•RESISTORS IN PARALLEL

•THE POTENTIAL DIVIDER

2 ©

John Parkinson

KIRCHHOFF’S FIRST LAWKIRCHHOFF’S FIRST LAW

THE ALGEBRAIC SUM OF THE CURRENTS

AT A JUNCTION IS ZERO

I1

I2

I3

I4

I5

I1 + I2 = I3 + I4 + I5

I = 0

©John Parkinson

3

PUTTING IT SIMPLY

EQUALS CURRENCURRENT OUTT OUT

CURRENT CURRENT ININ

Example 0.5 A

0.3 A

? A

0.5 A - 0.3 A = 0.2 A

©John Parkinson

4

KIRCHHOFF’S SECOND KIRCHHOFF’S SECOND LAWLAW = IR

AROUND A CLOSED CIRCUIT LOOP, THE ALGEBRAIC SUM OF THE e.m.f.s IS EQUAL TO THE ALGEBRAIC SUM OF THE p.d.s

I4

Then = IR1 + IR2 + (I-I4) R3

R1 R2 R3

R4

I

II-I4

IR1 IR2 (I-I4)R3

For R3 and R4 loop , 0 = I4R4 - (I-I4) with no e.m.f.’s in this loop

©John Parkinson

5

Example A battery charger has an e.m.f. of 14 V and an internal resistance of 1.6 . It is used to charge a 12 V battery, which itself has an internal resistance of 0.3 , through a 1 resistor. Find the charging current.

14 V

1.6

0.3 I

Charger

Charger

12 V

0.1

= 14V-12V = 2V

Sum of p.d.s = IR Ix0.3 + Ix0.1 +

Ix1.6

= Ix2So 2V = Ix2

I = 1A

©John Parkinson

6

RESISTORS IN SERIESRESISTORS IN SERIES

IR1 R2 R3

e.m.f.=

By Kirchhoff’s SECOND law = IR1 + IR2 + IR3

= I(R1+R2+R3)If we apply Ohm’s Law to the whole

circuit, we have = IR, if R is the total resistance

So IR = I(R1+R2+R3)

Dividing each side by I R = R1+R2+R3

©John Parkinson

7

RESISTORS IN PARALLELRESISTORS IN PARALLEL

R1

R2

R3

e.m.f. =

I I1

I2

I3

By Kirchoff’s FIRST law

I = I1 + I2 + I3We now apply Ohm’s Law to

each component and to the whole

circuit, letting R = the total resistance

321 RV

RV

RV

RV

DIVIDING EACH SIDE BY V

321

1111RRRR

©John Parkinson

8

Example Find the current in each resistor

A=6

B=12

C=6

Source = 20V

123

61

1211

R 4

312

R

So the total resistance in the

circuit, including A, is

4 + 6 = 10What is the current taken from the battery & the current in A?

The p.d. across B & C is given by V = IR = 2A X 4 = 8V

IB = V R = 8 12 = 2/3 Amps

IC = V R = 8 6 = 4/3 Amps

20V 10 = 2A

Hence find IB

8V

IB

and IC

IC

What is the combined resistance of B and C?

©John Parkinson

9

THE POTENTIAL DIVIDERA

B

C

I

Vin

Vout

R1

R2

If NO current is drawn

through the output, then

by Ohm’s Law

21 RRV

I in

and Vout = IxR2

so21

1

RR

RVV inout

The voltage is divided

in the same ratio a the resistors

©John Parkinson

10

Example

150V

Vout

10

k

5

k

(i) Find Vout

Volts

kkk

RRR

VV inout

100

51010

150

21

1

(ii) Find Vout if another

10 k resistor is connected across AB

A

B

102

101

1011

ABRSo RAB = 5 Ohms Vout = 75 V

WHY?