'n Battery met 'n onbekende emk (ε) en onbekende interne weerstand (r) word aan drie resistors, 'n hoë-weerstand-voltmeter, twee skakelaars en twee ammeters met weglaatbare weerstand verbind, soos hieronder getoon - NSC Physical Sciences - Question 8 - 2023 - Paper 1

To find the reading on the voltmeter when both switches are closed:

Using Ohm's Law and the known current ($I_1 = 1.5 A$) through the circuit:

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

We know:

• Total resistance $R_{total} = 3\,\Omega + 2\,\Omega + 5\,\Omega = 10\,\Omega$
• Hence, $V = I_1 imes R_{total} = 1.5 A imes 10\,\Omega = 15\,V$

When both switches are closed, the current divides through the circuit. The total current is 1.5 A and is shared between the parallel resistances:

Using the current ratios:

• The reading on ammeter A2 ( $\frac{3}{5}$) will be $I_2 = \frac{3}{8} \times 1.5 A = 0.5625 A$

Using the voltage across the 3 Ω resistor with current through it:

$P = I^2 R = (0.5625 A)^2 \times 3\,\Omega = 0.9453 W\approx 0.95 W$

When switch S2 is opened and switch S1 remains closed, the voltmeter reading will change due to the different circuit configuration. The potential difference will likely decrease since the circuit becomes less complex, and some resistances are no longer in the circuit:

• Since less current can flow through the resistor connected to the voltmeter, the reading will either decrease or remain the same, depending on the new configuration.