The battery in the circuit diagram below has an emf of 12 V and an internal resistance of 0,5 Ω - NSC Physical Sciences - Question 8 - 2018 - Paper 1

To find the reading on the voltmeter when switch S is open, we can apply Ohm's law. Since the ammeter reads 2 A with switch S open, the total resistance in the circuit can be calculated using:

$I = \frac{V}{R_{total}}$

Where:

• I = 2 A
• V = 12 V

Calculating: $R_{total} = \frac{12 V}{2 A} = 6 \Omega$

Now, knowing the internal resistance of the battery is 0.5 Ω, the effective resistance in the circuit (excluding R) with switch S open is:

$R_{effective} = R_{total} - r_{internal} = 6 \Omega - 0.5 \Omega = 5.5 \Omega$

The potential across the 2 Ω resistor can now be calculated by using Ohm's law on that resistor:

$V_{load} = I \times R = 2 A \times 2 \Omega = 4 V$

Thus the reading on the voltmeter is 4 V.

To find the resistance of resistor R, we can use the total voltage equation. Since there is 12 V and we have calculated a voltage drop of 4 V across the 2 Ω resistor:

The remaining voltage (the voltage across R) is:

$V_R = V_{battery} - V_{load} = 12 V - 4 V = 8 V$

We can use Ohm's law again to find R:

$R = \frac{V_R}{I} = \frac{8 V}{2 A} = 4 \Omega$

Thus, the resistance of resistor R is 4 Ω.

When switch S is closed, the total resistance in the circuit decreases due to R becoming part of the circuit. The current will increase as per Ohm's law, leading to a higher voltage drop across R. Consequently, the voltage reading on the voltmeter will decrease because the total emf (12 V) remains constant while the current through the circuit increases:

Therefore, the voltmeter reading will DECREASE as a result of closing switch S.