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The measurement of voltmeter in the following circuit is
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$2.25 \mathrm{~V}$
Resistance of voltmeter and $60 \Omega$ are in parallel.
$\therefore \quad$ Equivalent resistance
$=\frac{60 \times 40}{60+40}=\frac{60 \times 40}{100}=24 \Omega$
For the whole circuit, current $I=\frac{6 \mathrm{~V}}{(40+24) \Omega}=\frac{6}{64}=\frac{3}{32} \mathrm{~A}$
The voltmeter reads potential difference across $60 \Omega$.
$\begin{aligned} & \therefore \quad \text { Voltmeter reading }=I \times R_{e q} \\ & \qquad=\frac{3}{32} \times 24=\frac{9}{4} \mathrm{~V}=2.25 \mathrm{~V}\end{aligned}$
$\therefore \quad$ Equivalent resistance
$=\frac{60 \times 40}{60+40}=\frac{60 \times 40}{100}=24 \Omega$
For the whole circuit, current $I=\frac{6 \mathrm{~V}}{(40+24) \Omega}=\frac{6}{64}=\frac{3}{32} \mathrm{~A}$
The voltmeter reads potential difference across $60 \Omega$.
$\begin{aligned} & \therefore \quad \text { Voltmeter reading }=I \times R_{e q} \\ & \qquad=\frac{3}{32} \times 24=\frac{9}{4} \mathrm{~V}=2.25 \mathrm{~V}\end{aligned}$
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