Resistances $R_1$ and $R_2$ each $60\,\Omega$ are connected in series as shown in figure. The Potential difference between $A$ and $B$ is kept $120$ volt. Then what ............. $V$ will be the reading of voltmeter connected between the point $C$ and $D$ if resistance of voltmeter is $120\,\Omega .$
Medium
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According to the above problem,
the voltmeter with resistance $R_{v}=120$ ohm and resistor $R_{2}=60$
ohm are in parallel, hence equivalent resistance between them is
$R_{e}=\frac{R_{2} R_{v}}{R_{2}+R_{v}}$
$R_{e}=\frac{(60)(120)}{60+120}=40 \Omega$
Now the reading of voltmeter can be obtain by
$V_{v}=V\left(\frac{R_{e}}{R_{1}+R_{e}}\right)$
$V_{v}=120\left(\frac{40}{60+40}\right)$
$V_{v}=40\left(\frac{6}{5}\right)=48 \mathrm{V}$
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