Download Series and parallel resistors

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
I
But using Kirchoff's second rule the total potential difference
across them is V = V1 + V2
R1
R2
V1
V2
Therefore V = IR = IR1 + IR2 where R is the effective series
resistance of the two resistors.
So:
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
Figure 1
I = V/R = V/R1 + V/R2
R1
I
I1
A
I2
V
R2
Resistors in parallel:
1/R = 1/R1 + 1/R2
Figure 2
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!).
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Example Problems
1. Calculate the resistance of the following combinations:
(a) 100 Ω and 50 Ω in series
(b) 100 Ω and 50 Ω in parallel
(a) R = R1 + R2 = 100 + 50 = 150 
(b) 1/R = 1/R1 + 1/R2 = 1/100 + 1/50 = and so R = 33 
2. Calculate the current flowing through the following when a p.d of 12V is applied across the ends:
(a) 200 Ω and 1000 Ω in series
(b) 200 Ω and 1000 Ω in parallel
(a) resistance = 1200  Using I = V/R = 12/1200 = 0.01 A = 10 mA
(b) resistance = 167 . Using I = V/R = 12/167 = 0.072 A = 72 mA
3. You are given one 100 Ω resistor and two 50 Ω resistors. How would you connect any combination of
them to give a combined resistance of: (a) 200 Ω (b) 125 Ω
(a) 100  in series with both the 50  .
(b) the two 50  in parallel and this in series with the 100 
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 and a 100 resistor connected first in (a) series and
then (b) in parallel
(a) Series resistance = 50 + 100 = 150 
(b) Parallel resistance = 50x100/[50+100] = 5000/150 = 33.3 
2. Calculate the combined resistance of two 50 resistors connected in parallel, the combination being
joined in series to a further 50 resistor.
Parallel section resistance = 25 
Then in series with a further 50  giving a total resistance of 75 .
A further note on circuits
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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
B
D
difference across the 5 Ω resistor = (5/10)x12 = 6V
6
Therefore potential at A = 6V
Potential drop through the 6Ω resistor = (6/9)x12 = 8V
5
3
Therefore potential at B = 4V
12V
And so the potential difference between A and B = 6 - 4 = 2V
A
5
C
Figure 3
Some other interesting resistor networks may be found in the following file:
See: Advanced text/Electricity and magnetism/Current electricity/Resistance networks
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