The battery in the circuit shown below has an emf of 12 V and an unknown internal resistance r - NSC Physical Sciences - Question 8 - 2021 - Paper 1

With switch S open, voltmeter V₂ also shows 0 V since there is no potential difference across it.

Power is defined as the rate at which work is done or energy is transferred. It can be expressed mathematically as:
P = rac{W}{t}
where P is power, W is work done, and t is the time taken.

Using the formula for power,
$P = I^2 R$
we rearrange it to find resistance:
R = rac{P}{I^2}
Substituting the known values:
R_X = rac{5.76 ext{ W}}{(1.2 ext{ A})^2} = 4 ext{ Ω}.

The total external resistance (R_total) in the circuit is the sum of the resistance of resistor X and the additional external resistances. In this case, it is just the resistance of X, which is 4 Ω.

To find the reading on voltmeter V₂, apply Ohm’s law:
$V = IR$
We can find the voltage across the 2.4 Ω resistor:
$I = 1.2 ext{ A}, R = 2.4 ext{ Ω}$
Thus,
$V_2 = 1.2 imes 2.4 = 2.88 ext{ V}$.

When the switch is closed and the wire of negligible resistance connects points P and Q, the total resistance in the circuit decreases.
For V₁, the reading will DECREASE because the current increases, leading to a greater voltage drop across the internal resistance of the battery.