8.1 In the circuit below the battery has an emf (ϵ) of 12 V and an internal resistance of 0.2 Ω - NSC Physical Sciences - Question 8 - 2016 - Paper 1

When switch S is open, the voltmeter is connected across points a and b, where there is no current flowing through the circuit. Thus, the reading on the voltmeter will equal the emf of the battery, which is 12 V.

When switch S is closed, the voltmeter is connected across points c and d, where the resistance of the circuit is present. The potential difference measured by the voltmeter is given as 11.7 V.

To calculate the current in the battery, we use Ohm's law, which states that the current (I) is given by the formula:

$I = \frac{V}{R}$

In this case, the voltage across the battery is 11.7 V and the total resistance in the circuit includes the internal resistance of the battery (0.2 Ω) and the resistor R.

Let R be the equivalent resistance in the circuit.

Using the information that the total current I is 1.5 A, we calculate as follows:

$11.7 = I_{total} \cdot (R + 0.2)$

This further simplifies to find the current, and substituting gets us the value of 1.5 A.

To find the effective resistance of the parallel branch involving resistors, we can use the formula:

$\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$

Here R1 and R2 are the resistances (10 Ω and 15 Ω respectively). Therefore:

$\frac{1}{R_{eff}} = \frac{1}{10} + \frac{1}{15}$

Calculating gives:

$R_{eff} = 6 \Omega$.

From the earlier calculations, we derived the total current flowing and using the effective voltage across the internal resistor.

Using the relationship:

$V = I \cdot R$, where V is known (11.7 V), the internal resistance (0.2 Ω) can be taken into account. Solving for R gives:

1.8 Ω.

Therefore, the resistance of resistor R is 1.8 Ω.