Series and Parallel Circuits



Components in a circuit can be connected in series or parallel.
A series arrangement of components is where they are inline with each other, i.e. connected
end-to-end.
A parallel arrangement of components is where they are connected across each other where the
current has more than one path through that part of the circuit.
Measuring Current and Potential Difference or Voltage


Electric current is measured using an ammeter which is connected in series with the component.
Potential difference (p.d.), or voltage, is measured using a voltmeter which is connected in
parallel with the component.
Current and Voltage (p.d.) – Series Circuits


The current is the same at all points in a series circuit.
The sum of the potential differences across the components in a series circuit is equal to the
voltage of the supply.
Current and voltage (p.d) – Parallel Circuits


The potential difference across components in parallel is the same for all components.
The sum of the currents in parallel branches is equal to the current drawn from the supply.
Resistors in Series

When more than one component is connected in series, the total resistance of all the
components is equivalent to one single resistor, RT, calculated using the relationship
Resistors in Parallel

When more than one component is connected in parallel, the total resistance of all the
components is equivalent to one single resistor, RT, calculated using the relationship
Example – mixed series and parallel circuit
Calculate the total resistance of this circuit
1) Work out which resistors are in series and which are in parallel
 Simplify one step at a time
2) Deal with series combination R2 and R3 first
RT = R2 + R3 = 4 + 8 = 12 Ω
3) Simplified version looks like this


The combined resistance, RA of 12 Ω is in parallel with resistor R4
This combination is in series with resistor R1
=
which means
4) This is the same as
RT = R1 + Rcomb = 6 + 6 = 12 Ω

A single resistance of 12 Ω is equivalent to the 4 resistor combination.
Advantages of parallel circuits
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

If one component stops working the rest keep working
Can have combinations of components switched on and off as desired
All components connected in parallel will have the same voltage across them
o Connecting components to the mains supply of 230 V means that all components
receive the full 230 V
Potential divider circuits

A potential divider is a device or a circuit that uses two (or more) resistors in series, or a
variable resistor (potentiometer) to
o provide a fraction of the available voltage (p.d.) from the supply.


The supply voltage is shared (divided up) between the components
The larger resistance takes the larger share of the supply voltage in direct proportion
o The ratio of resistances = the ratio of voltage shared between the resistors
Example 2
Consider the following voltage divider circuit:
20 kΩ
Y
12 V
10 kΩ
X
(a)
Calculate the p.d. across X Y.
V1 
R 1 = 10 kΩ
R1
 VS
R1  R2
R 2 = 20 kΩ
V S = 12 V
10
 12
10  20
V1  4 V
V1 
V1 = ?
(b) Another 10 k resistor is added as shown. Calculate
V 1 = 4the
V new p.d. across X Y.
20 kΩ
12 V
Y
10 kΩ
10 kΩ
X
R 1 = 5 kΩ
R 2 = 20 kΩ
V S = 12 V
V1 = ?
V1 
R1
 VS
R1  R2
5
 12
5  20
V1  2.4 V
V1 
V 1 = 2.4 V
Electrical Energy and Power
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
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When there is current in a component, an energy change takes place.
o e.g. when there is current in the filament in a lamp, the electrical energy is converted to
heat and light.
o In an electrical heating element, the energy change takes place in the resistance wire
Electrical energy, like all types of energy is measured in Joules.
Power is the rate of energy transfer (same definition as on page 4 for mechanical power)
Energy (J)
Power (W)
time (s)

The units of power are Watts (W). 1 Watt means 1 Joule of energy is transferred every second.
Power, current and voltage

The power (P) is also equal to the current (I) in Amps multiplied by the voltage (V) in volts.
voltage (V)
power (W)
current (A)
Example
An electric heater operates on the mains voltage of 230 V. The heating elements operate at a
current of 8 A.
Calculate the power developed by the heater.
V = 230 V
I=8A
P=?
P = IV = 8 x 230 = 1840 W
More Power equations

The equations P = IV and V = IR can be combined by substitution to form two other equations.
P = I2R
P = IV and substitute V = IR into the equation for V gives P = I  IR = I2R
P = V2/R
P = IV and substitute I =

into the equation for I gives P =
xV=
This means there are three expressions for the power which are equivalent to each other
Example 1
An electric heater operates on the mains voltage of 230 V. The heater has a power rating of 2∙0 kW.
Calculate the resistance of the heating element of the heater.
P = 2 kW
V = 230 V
R=?
P=
R=
Example 2
In the circuit shown opposite calculate:
40 V
(a)
the current in each resistor
(b)
the p.d. across each resistor
(c)
the power dissipated in each resistor.
4Ω
6Ω
5Ω
10 Ω
(a)
R S = R 1 + R 2 = 4 + 6 = 10 Ω
1
1 1 1
  
Rp 10 10 5
RP = 5 Ω
R T = 5 + 5 = 10 Ω
IT 
VT 40

 4 A (current through 5 Ω)
RT 10
The resistance of each parallel section is 10 
so current through 6 , 4  and 10  resistors all = 2 A
(b)
For the 5 Ω resistor: V = IR = 4 × 5 = 20 V
for the 4 Ω resistor: V = IR = 2 × 4 = 8 V
for the 6 Ω resistor: V = IR = 2 × 6 = 12 V
for the 10 Ω resistor: V = IR = 2 × 10 = 20 V
(c)
For 5 Ω: P = I 2 R = 4 2 × 5 = 80 W
for 4 Ω: P = I 2 R = 2 2 × 4 = 16 W
for 6 Ω: P = I 2 R = 2 2 × 6 = 24 W
for 10 Ω: P = I 2 R = 2 2 × 10 = 40 W