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Q. Figure shows a potentiometer. Length of the potentiometer wire $ AB $ is $ 100 \,cm $ and its resistance is $ 100\, \Omega $ . $ EMF $ of the battery $ E $ is $ 2 V $ . A resistance $ R $ of $ 50 \Omega $ draws current from the potentiometer. What is the voltage across $ R $ when the sliding contact $ C $ is at the mid-point of $ AB $ ?
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AMUAMU 2018Current Electricity

Solution:

Given, $E = 2 V$, $R_{AB}=100\,\Omega$
$l_{AB}=100\,cm$ and $R=50\,\Omega$
Circuit according to the question,
image
$AC=CB=\frac{l_{AB}}{2}=50\,cm$
$R_{AB}=100\,\Omega$
$R_{AC}=R_{CB}=\frac{R_{AB}}{2}=\frac{100}{2}$
$\therefore R_{AC}=R_{CB}=50\,\Omega$
$\therefore $ Net resistance of the circuit, $R_{net}=R_{CB}+\frac{R\cdot R_{AC}}{R+R_{AC}}$
$=50+\frac{50\times50}{50+50}=50+25$
$R_{net}=75\,\Omega$
$\therefore $ Current from the battery, $I=\frac{E}{R_{net}}=\frac{2}{75}A$
Using $KVL$ in mesh $1$,
$E-V_{AC}-V_{CB}=0$
$\Rightarrow V_{AC}+V_{CB}=2$
$\Rightarrow V_{AC}+IR_{CB}=2$
$\Rightarrow V_{AC}=2-\frac{2}{75}\times50$
$\Rightarrow V_{AC}=2-\frac{4}{3}=\frac{6-4}{3}=\frac{2}{3}V$