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Series and parallel resistors

In this section we deal with the mathematics of more than one resistor in a series or parallel circuit.

Two resistors in series (Figure 1)

The current (I) flowing through R1 and R2 is the same and so the potential differences across them are V1 = IR1 and V2 = IR2

But using Kirchoff's second rule the total potential difference across them is
V = V1 + V2

Therefore V = IR = IR1 + IR2 where R is the effective series resistance of the two resistors.

Resistors in series:  R = R1 + R2

Two resistors in parallel (Figure 2)

The potential difference (V) across each of the two resistors is the same, and the current (I) flowing into junction A is equal to the sum of the currents in the two branches (Kirchoff's first rule) therefore:
I = I1 + I2

But since V = I1R1 = I2R2 I = V/R = V/R1 + V/R2

Resistors in parallel:  1/R = 1/R1 + 1/R2

where R is the effective resistance of the two resistors in parallel.

Notice that two resistors in series always have a larger effective resistance than either of the two resistors on their own, while two in parallel always have a lower resistance. This means that connecting two or more resistors in parallel, such as the use of a mains adaptor, will increase the current drawn from a supply (Look at the section that deals with the bath with the two plug holes!).

Example problems
1. Calculate the resistance of the following combinations:
(a) 100 W and 50 W in series
(b) 100 W and 50 W in parallel
(a) R = R1 + R2 = 100 + 50 = 150 W
(b) 1/R = 1/R1 + 1/R2 = 1/100 + 1/50 = and so R = 33 W

2. Calculate the current flowing through the following when a p.d of 12V is applied across the ends: (a) 200 W and 1000 W in series
(b) 200 W and 1000 W in parallel
(a) resistance = 1200 W. Using I = V/R = 12/1200 = 0.01 A = 10 mA
(b) resistance = 167 W. Using I = V/R = 12/167 = 0.072 A = 72 mA

3. You are given one 100 W resistor and two 50 W resistors. How would you connect any combination of them to give a combined resistance of: (a) 200 W (b) 125 W
(a) 100 W in series with both the 50 W .
(b) the two 50 W in parallel and this in series with the 100 W

For three resistors in series the combined resistance is:

R = R1 + R2 + R3

and for three resistors in parallel it is:

1/R = 1/R1 + 1/R2 + 1/R3

Resistors in parallel an alterative formula

The formula for two resistors in parallel may also be written as:

R = R1R2/[R1 + R2]

N.B this version cannot be extended simply to cover the case of three resistors in parallel.

The version for three resistors being: R = R1R2R3/[R1R2 + R1R3 + R2R3]

Example problems
1. Calculate the combined resistance of a 50 W and a 100 W resistor connected first in (a) series and then (b) in parallel
(a) Series resistance = 50 + 100 = 150 W
(b) Parallel resistance = 50x100/[50+100] = 5000/150 = 33.3 W

2. Calculate the combined resistance of two 50 W resistors connected in parallel, the combination being joined in series to a further 50 W resistor.

Parallel section resistance = 25 W
Then in series with a further 50 W giving a total resistance of 75 W.

A further note on circuits

The p.d. between the points A and B in the circuit in Figure 3 may be found by considering the ratio of the voltage drops in the resistors in each branch of the circuit. If the potential at C is zero then:
Since the total resistance between D and C is 10? the potential difference across the 5 O resistor = (5/10)x12 = 6V
Therefore potential at A = 6V
Potential drop through the 6O resistor = (6/9)x12 = 8V
Therefore potential at B = 4V
And so the potential difference between A and B = 6 - 4 = 2V

Some other interesting resistor networks may be found in the following file:
Resistance networks
© Keith Gibbs 2011