Fig. 5.1 illustrates a circuit that includes a battery, two fixed resistors X and Y, and a light-dependent resistor (LDR) Z. The battery has electromotive force (e.m.f.) $5.0\,\text{V}$ and internal resistance $4.7\,\Omega$. The current through X is $I_1$ and the current through Y is $I_2$. The resistance of X is $100\,\Omega$. The resistance of Z changes with the intensity of light falling on it, as shown in Fig. 5.2.
(a)[1]
State Kirchhoff’s first law
(b(i))[2]
The light intensity incident on Z is $130\,\text{W m}^{-2}$. The current in the battery is $38\,\text{mA}$.
Show that the terminal potential difference of the battery is $4.8\,\text{V}$.
(b(ii))[3]
Calculate the current $I_2$ in Y
(b(iii))[2]
Calculate the power dissipated by Y
(b(iv))[3]
The light intensity incident on Z decreases.
State and explain how this changes the terminal potential difference of the battery.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “sum of current(s) entering a junction = sum of current(s) leaving junction or (algebraic) sum of current(s) at a junction is zero” …