The E ect of Internal Resistance on Practical Circuits

1.

Answers
Questions - The E ect of Internal Resistance on Practical Circuits
1.
a)
Electromotive force is when the charges gain electrical energy from a power supply. It is defined as the energy
transferred per unit charge. It has units of Volts, V.
b)
Potential Di erence is when the charges lose electrical energy to a component. It is also defined as the energy
transferred per unit charge and has units of Volts, V.
c)
The terminal potential di erence is the potential di erence across the terminals of a battery, after “lost volts”
have been taken into account due to the internal resistance.
2.
a)
E = V + Ir = IR + Ir = I(R + r)
I=
3.
E
6
=
= 0.35A
R+r
17
b)
V = IR = 0.35x15 = 5.29V = 5.3V
c)
6.0 − 5.3 = 0.70V
d)
I=
a)
Ammeter reading: I =
b)
Voltmeter reading: V = IR = 2.67x 3 = 8V
c)
New ammeter reading: I =
E
6.0
=
= 3.0A
r
2
E
12
=
= 2.67A
R+r
4.5
E
12.0
=
= 4.0A
r
3
New voltmeter reading: V = IR = 4x1.5 = 6V
4.
I=
V
1.0
=
= 0.05A
R
20
1.0 = 1.5 − (0.05xr), therefore r = 10Ω

2.

Answers Continued
Questions - The E ect of Internal Resistance on Practical Circuits
5.
a)
I=
E
600
=
= 13.95A
R+r
43
b)
I=
E
600
=
= 200A
r
3
c)
To decrease the short circuit current and hence increase safety. It will also lessen the damage caused to the
supply.
6.
A car battery has a very low internal resistance around 3m Ω .
When you start the motor, the battery needs to deliver a large amount of power to turn the engine. This would
typically require it to deliver an enormous current of around 400 A
The battery voltage is normally 12V, but when you start the engine, the current of 400A (or more) through the internal
resistance of 3m Ω means there is a voltage drop at the terminals of Ir = 400x (3x10−3) = 1.2V
This means that the battery's operating "terminal" voltage drops briefly to 12 - 1.2= 10.8V
This is why the lights dim, as they operate at 12V normally, but suddenly find they only have 10.8V.
7.
Overleaf

3.

Answers Continued
Questions - The E ect of Internal Resistance on Practical Circuits
7.
a)
R(Ω)
I (A)
V (V)
P(W)
0.0
7.50
0.00
0.00
0.4
5.00
2.00
10.00
0.8
3.75
3.00
11.25
1.2
3.00
3.60
10.80
1.6
2.50
4.00
10.00
2.0
2.14
4.29
9.18
b)
c)
Internal Resistance = gradient of the graph: r = 0.8Ω
emf = y-intercept of the graph: E = 6.0V

4.

Answers Continued
Questions - The E ect of Internal Resistance on Practical Circuits
7.
Continued
d)
.
I.
Power initially increases with resistance, peaks at around 0.8-0.9Ω and then decreases. NOTE: the
mathematical relationship between power of the component and resistance can be easily proved to be:
E2R
, which translates to the following graph and matches the experimental graph perfectly:
P=
(r + R)2
II.
From the graph, maximum power is at 0.8Ω. Theoretically, the maximum power output of the external
resistor happens when R = r