A battery with electromotive force (e.m.f.) $12\,\text{V}$ and negligible internal resistance is linked to a circuit containing two lamps and two resistors, as illustrated in Fig. 6.1. The two lamps in the circuit have the same resistance. The two resistors have resistances $R$ and $28\,\Omega$. The lamps are joined at junction X and the resistors are joined at junction Y. The current in the battery is $0.50\,\text{A}$ and the current in the lamps is $0.20\,\text{A}$.
(a(i))[2]
Calculate the resistance of each lamp.
(a(ii))[2]
Calculate resistance $R$.
(b)[3]
Determine the potential difference $V_{XY}$ between points X and Y.
(c)[2]
Calculate the ratio
$\dfrac{\text{total power dissipated by the lamps}}{\text{total power supplied by the battery}}$.
(d)[2]
The resistor of resistance $R$ is now replaced by another resistor of lower resistance. State and explain the effect, if any, of this change on the ratio in (c).
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use of $R = V/I$” …