The circuit diagram below shows a battery with an emf of 12 V and an internal resistance of 0.5 Ω connected to three resistors, a light bulb, a switch, an ammeter and connecting wires - NSC Physical Sciences - Question 8 - 2023 - Paper 1

To find the total external resistance of the circuit, we first identify the configuration of resistors. Resistors R1 and R2 are in parallel, thus:

$R_{12} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 10}{10 + 10} = 5 \Omega$

Now, we find the total resistance from the combination of R12 in series with R3:

$R_{total} = R_{12} + R_3 = 5 + 15 = 20 \Omega$

Using Ohm's law, we find the total current flowing in the circuit. The total voltage across the total resistance (including internal resistance) is:

$V_{total} = 12V$

The effective resistance including internal resistance is:

$R_{effective} = R_{total} + r = 20 + 0.5 = 20.5 \Omega$

Thus, the current (I) is:

$I = \frac{V_{total}}{R_{effective}} = \frac{12}{20.5} \approx 0.585 A$

To find the power dissipated by resistor R3, we first need the voltage across it, which can be calculated from the current:

Using the total voltage across the complete circuit:

$V_{ext} = I \times R_{total} = 0.585 \times 20 = 11.7V$

The voltage across R3 can be calculated using the current through R3 (same as total current):

$V_R3 = I \times R_3 = 0.585 \times 15 = 8.775V$

Now, the power (P) through the resistor R3 can be calculated using:

$P = I^2 \times R = (0.585)^2 \times 15 \approx 5.14W$

DECREASES

When switch S is opened, the total resistance in the circuit increases because the light bulb is no longer in a parallel configuration with the resistors R1 and R2. This increase in total resistance leads to a reduction in the total current flowing through the circuit, which results in a decrease in power dissipation across the light bulb. Consequently, this decreases its brightness.