Edexcel A-Level Physics Electric Circuits

Use $A=\pi r^2$ and $R=\rho l/A$.
$$A=\pi(0.0905\times10^{-3})^2=2.57\times10^{-8}\,\mathrm{m^2},$$$$R=\frac{(1.10\times10^{-6})(0.550)}{2.57\times10^{-8}}=23.5\,\Omega.$$

Maximum power occurs at minimum total resistance. With X and Y closed, $R_{tot}=(1/47+1/23.5)^{-1}=15.7\,\Omega$. Thus $P=V^2/R=12.0^2/15.7=9.2\,\mathrm{W}$, less than $12\,\mathrm{W}$, so the suggestion is incorrect.

ai

$$\text { Use of } \frac{1}{R}=\frac{1}{500 \Omega}+\frac{1}{600 \Omega}$$

Use of V=I R

Voltmeter reading $=6.3(\mathrm{~V})$

Example of calculation
$$\begin{aligned} & \frac{1}{R_{\text {parallel }}}=\frac{1}{500 \Omega}+\frac{1}{600 \Omega}, \text { so } R_{\text {parallel }}=273 \Omega & V=I R=23.0 \times 10^{-3} \mathrm{~A} \times 273 \Omega=6.28 \mathrm{~V} \end{aligned}$$

aii

p.d. across thermistor $=12 \mathrm{~V}-6.3 \mathrm{~V}$ (allow ecf from ai)

Use of P=V I (allow use of $P=I^{2} R$ or $P=V^{2} / R$ )
$$P=0.13 \mathrm{~W}$$

Example of calculation

p.d. across thermistor $=12.0 \mathrm{~V}-6.3 \mathrm{~V}=5.7 \mathrm{~V}$$$\begin{aligned} & P=V I=5.7 \mathrm{~V} \times 23 \times 10^{-3} \mathrm{~A} & P=0.13 \mathrm{~W} \end{aligned}$$

Resistance (of circuit/thermistor) increases

Ammeter reading decreases (dependent on MP1)

Or Current decreases (dependent on MP1)

p.d. across thermistor increases

Or V=I R for fixed/parallel resistors, and I decreases

Reading on voltmeter decreases (dependent on MP3)

Or p.d. across fixed/parallel resistors decreases (dependent on MP3)

(Allow 1 mark maximum if stated that both ammeter and voltmeter readings decrease, and no other marks have been awarded)