1. Series and parallel circuits
Resistors

2. This network shows a circuit which has both
a series circuit and a parallel circuit.
Parallel (10 + 20 + 30) + Series (10)
20Ω
30Ω
10Ω
110V
10Ω

3. To work out the total resistance in the
network you must first work out the
resistance in the parallel circuit.

4. Once the amount of resistance in the
parallel circuit has been found the
next step is to work out the value of the
resistances in series. The total value
of resistance in parallel is added to the
resistors in series. This will give you the
total value of resistance for the whole
network.

5. (1) Find the parallel resistance
1 = 1 + 1 + 1
Rp R1 R2 R3
1 = 1 + 1 + 1
Rp 10 20 30
1 = 6 + 3 + 2
Rp 60
1 = 11
Rp 60
Rp = 60
1 11
So Rp = 5.45Ω

6. (2) Now add the equivalent resistor to the
series resistor.
Rt = Rp + Rs
Rt = 5.45 + 10
Rt = 15.45Ω

7. (3) Find the current using Ohm’s Law.
I = V
R
I = 110
15.45
I = 7.12A

8. Voltage drop
Circuit conductors are like
resistors, in that the longer the
conductor, then the higher the resistance
and the greater the voltage drop.

9. Using Ohm’s Law you can work out
the circuit’s actual voltage drop.
BS7671 has tables of manufacturers’
data about voltage drop.
The voltage drop is listed in terms of
milli-volts per amp per metre (mV/A/m).
The voltage drops are for conductor feed
and return, eg for two single core cables or for
one two-core cable.

10. The wiring regulations (BS7671) say
that voltage drop between the origin of
the installation (supply terminals) and the
current-using equipment shall not exceed
4% of the nominal voltage of supply, eg
9.2V for a supply voltage of 230V.

11. A radial circuit is wired in 50mm copper cable. If the
voltage drop for this conductor is 0.95 mV/A/m find:
a. the current in each section;
b. the volt drop in each section;
c. the supply voltage.
A B C
50 M 30 M
40A 55A

12. Find the current that flows in section A-B
and the circuit length in section A-B.
Current in A-B = 40A + 55A = 95A

13. Work out the voltage drop in A-B by
using:
Volt drop = mV/A/m x I x length
1000

14. VD of A-B = 0.95 x 95 x 50
1000
VD = 4.51 volts

15. Current flow in B-C and circuit length B-C.
Current B-C = 55A
Length of section B-C = 30m
VD(B-C) = 0.95 x 55 x 30
1000
VD = 1.56V

16. The voltage drop total
4.51 + 1.56 = 6.08
The supply voltage is
200V + 6.08V = 206.08V