Question 82 of 380

The actual value of resistance $R$ , shown in the figure is $30\Omega .$ This is measured in an experiment as shown using the standard formula $R=\dfrac{V}{I}$ , where $V$ and $I$ are the readings of the voltmeter and ammeter, respectively. If the measured value of $R$ is $5$ less, then the internal resistance of the voltmeter is:

A$35\Omega$B$600\Omega$C$570\Omega$D$350\Omega$

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Answer: (C) $570\Omega$

Measured value of $R=5$ less than
actual value of $R$ i.e., ${R}^{'}=30-(5$ of $30$$)=30-\dfrac{5}{100}\times 30$
${R}^{'}=28.5\Omega$ $...\left(i\right)$

Now, let us assume that internal resistance of voltmeter ${R}_{V}$. Replacing voltmeter with its internal resistance, we get following circuit.
It is clear that the measured value, ${R}^{'}$ should be equal to parallel combination of $R$ and ${R}_{V}$. Mathematically,
${R}^{'}=\dfrac{R\times {R}_{V}}{R+{R}_{V}}=28.5\Omega$
Given. $R=30\Omega$
$\Rightarrow \dfrac{30{R}_{V}}{30+{R}_{V}}=28.5$
$\Rightarrow 30{R}_{V}=\left(28.5\times 30\right)+28.5{R}_{V}$
$\Rightarrow 1.5{R}_{V}=28.5\times 30$
$\Rightarrow {R}_{V}=\dfrac{28.5\times 30}{1.5}=19\times 30$
or ${R}_{V}=570\Omega$