In the diagram below, a battery with an emf of 12 V and an internal resistance of 1 Ω, is
connected to three resistors with resistance R, 2R and R. A voltmeter V is connected across the
battery. The ammeter A has a negligible resistance.
S1 and S2 are now closed and a current of 2,4 A is registered on A.
4.1 Calculate the total resistance of the external circuit in terms of R.
4.2 Calculate the value of R.
4.3 Calculate the reading of V.
4.4 S1 is now opened and the readings of A and V taken. Say what will happen to each of the
readings. Just write down DECREASE, INCREASE or SAME
4.4.1 A:
4.4.2 V:
4.1 Total resistance of the external circuit can be found by adding the resistances of the three resistors:
Total resistance = R + 2R + R = 4R
So, total resistance in terms of R is 4R.
4.2 To find the value of R, use Ohm's Law:
Total current in the circuit = 2.4 A
Total resistance = 4R
Voltage of battery = EMF - Ir_internal = 12 V - (2.4 A)(1 Ω) = 12 V - 2.4 V = 9.6 V
Using Ohm's Law: V = IR
9.6 V = 2.4 A * R
R = 9.6 V / 2.4 A
R = 4 Ω
So, the value of R is 4 Ω.
4.3 The reading of V can be found using the total resistance (4R = 4*4 = 16 Ω) and the total current (2.4 A):
V = IR
V = (2.4 A)(16 Ω)
V = 38.4 V
Therefore, the reading of V is 38.4 V.
4.4
4.4.1 A: DECREASE
4.4.2 V: INCREASE
When S1 is opened, the total resistance of the circuit will decrease because one of the resistors (R) is disconnected. This will result in an increase in current (A) flowing through the circuit. The voltage reading (V) will increase as well because the internal resistance of the battery is no longer in the circuit affecting the total voltage.