Series and parallel resistors - schoolphysics



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.

So:

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

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!).

For three resistors in series the combined resistance is:

and for three resistors in parallel it is:

Resistors in parallel – an alterative formula

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

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]

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 10Ω the potential difference across the 5 Ω resistor = (5/10)x12 = 6V

Therefore potential at A = 6V

Potential drop through the 6Ω 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:

See: Advanced text/Electricity and magnetism/Current electricity/Resistance networks

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V1 V2

Figure 1

R1

R2

I

Resistors in series: R = R1 + R2

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

Example Problems

1. Calculate the resistance of the following combinations:

(a) 100 © and 50 © in series

(b) 100 © and 50 © in parallel

(a) R = R1 + R2 = 100 + 50 = 150 Wð

(b) 1/R = 1/R1 + Ω and 50 Ω in series

(b) 100 Ω and 50 Ω in parallel

(a) R = R1 + R2 = 100 + 50 = 150 Ω

(b) 1/R = 1/R1 + 1/R2 = 1/100 + 1/50 = and so R = 33 Ω

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 Ω. Using I = V/R = 12/1200 = 0.01 A = 10 mA

(b) resistance = 167 Ω. 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 Ω in series with both the 50 Ω .

(b) the two 50 Ω in parallel and this in series with the 100 Ω

R = R1 + R2 + R3

R = R1R2/[R1 + R2]

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

Example problems

1. Calculate the combined resistance of a 50Ω and a 100Ω resistor connected first in (a) series and then (b) in parallel

(a) Series resistance = 50 + 100 = 150 Ω

(b) Parallel resistance = 50x100/[50+100] = 5000/150 = 33.3 Ω

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

Parallel section resistance = 25 Ω

Then in series with a further 50 Ω giving a total resistance of 75 Ω.

I1

I2

I

V

R2

R1

A

Figure 2

D

C

3Ω

6Ω

5Ω

5Ω

12V

B

A

Figure 3

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