1. KIRCHHOFF’S LAWS & COMBINATIONS OF RESISTORS
RESISTORS IN SERIES
RESISTORS IN PARALLEL
THE POTENTIAL DIVIDER
John Parkinson

2. THE ALGEBRAIC SUM OF THE CURRENTS
KIRCHHOFF’S FIRST LAW
I = 0
THE ALGEBRAIC SUM OF THE CURRENTS
AT A JUNCTION IS ZERO I3 I4 I1 I2 I5
I1 + I2 = I3 + I4 + I5
John Parkinson

3. PUTTING IT SIMPLY EQUALS CURRENT OUT
CURRENT IN
EQUALS
CURRENT OUT
0.5 A
Example
? A
0.3 A
0.5 A A = 0.2 A
John Parkinson

4. KIRCHHOFF’S SECOND LAW
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  R1 R2 R3 R4 I
IR1
IR2
(I-I4)R3 I
I-I4 I4
Then  = 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.
Charger
 = 14V-12V = 2V
0.1 
1.6 
Sum of p.d.s =  IR
Ix0.3 + Ix0.1 + Ix1.6
= Ix2
12 V
14 V
So 2V = Ix2
0.3 
Charger
I = 1A I
John Parkinson

6. By Kirchhoff’s SECOND law
RESISTORS IN SERIES
e.m.f.=  R1 R2 R3 I
 = IR1 + IR2 + IR3
= I(R1+R2+R3)
By Kirchhoff’s SECOND law
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)
R = R1+R2+R3
Dividing each side by I
John Parkinson

7. By Kirchoff’s FIRST law
RESISTORS IN PARALLEL
e.m.f. = 
By Kirchoff’s FIRST law 
I = I1 + I2 + I3 I I1 R1
We now apply Ohm’s Law to
each component and to the whole
circuit, letting R = the total resistance I2 R2 I3 R3
DIVIDING EACH SIDE BY V
John Parkinson

8. Example Find the current in each resistor Source = 20V
What is the combined
resistance of B and C?
Source = 20V IB
B=12
So the total resistance in the
circuit, including A, is
4 + 6  = 10
A=6 8V
C=6
What is the current taken from the battery & the current in A? IC
20V  10  = 2A
The p.d. across B & C is given by V = IR = 2A X 4  = 8V
Hence find IB
and IC
IB = V  R = 8  12 = 2/3 Amps
IC = V  R = 8  6 = 4/3 Amps
John Parkinson

9. in the same ratio a the resistors
THE POTENTIAL DIVIDER A
If NO current is drawn
through the output, then
by Ohm’s Law I R1
Vin B R2
Vout
and Vout = IxR2 C so
The voltage is divided
in the same ratio a the resistors
John Parkinson

10. Example 150V A Vout B So RAB = 5 Ohms Vout = 75 V WHY? (i) Find Voutk A 10 k
Vout
(ii) Find Vout if another
10 k resistor is connected across AB B
So RAB = 5 Ohms
Vout = 75 V
WHY?
John Parkinson