Paper 2 DC Circuits Answers

Q1 Series and parallel network

1. $1/R_p=1/6.0+1/3.0=0.500$, so $R_p=2.0\Omega$. [2]
2. Total $R=4.0+2.0=6.0\Omega$, so $I=12.0/6.0=2.0\mathrm{A}$. [2]
3. $V_4=IR=2.0(4.0)=8.0\mathrm{V}$. [2]
4. Parallel pair p.d. $=12.0-8.0=4.0\mathrm{V}$. Current in $3.0\Omega$ branch $=4.0/3.0=1.33\mathrm{A}$. [2]

Q2 Potential divider sensor

1. $V_{out}=6.0(3.0)/(2.0+3.0)=3.6\mathrm{V}$. [2]
2. For an NTC thermistor, resistance decreases as temperature increases. Since output is across the lower thermistor, $V_{out}$ decreases. [2]
3. The output is the fraction of supply across the component where it is measured. Moving the sensor changes whether decreasing sensor resistance decreases or increases the measured output. [2]

Q3 Potentiometer

1. Potential gradient $=2.00/1.00=2.00\mathrm{V}{\mathrm{m}}^{-1}$. [1]
2. $\mathcal{E}=2.00(0.720)=1.44\mathrm{V}$. [2]
3. At balance the detector current is zero, so the cell is not supplying current through the detector branch and its terminal p.d. equals its emf approximately. [2]