A battery with electromotive force $12\,\text{V}$ and negligible internal resistance is linked to two resistors and a light-dependent resistor (LDR), as shown in Fig. 4.1. An ammeter is joined in series with the battery. The LDR and switch S are connected across points X and Y.
(a)[3]
With the switch S open, Calculate the potential difference (p.d.) across XY.
(b)[3]
With the switch S closed, and the LDR resistance equal to $4.0\,\text{k}\Omega$, Calculate the current in the ammeter.
(c(i))[2]
The switch $S$ remains closed. The intensity of the light on the LDR is increased. State and explain the change in the ammeter reading.
(c(ii))[2]
The switch $S$ remains closed. The intensity of the light on the LDR is increased. State and explain the change in the p.d. across $XY$.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “The total resistance is $20\,\text{k}\Omega$, so the current is $12/20 = 20\,\text{mA}$.” …