Current Electricity 5 Question 3
3. 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 $R=\frac{V}{I}$, 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
(2019 Main, 10 Jan II))
(a) $600 \Omega$
(b) $570 \Omega$
(c) $350 \Omega$
(d) $35 \Omega$
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Solution:
- Measured value of $R=5 %$ less than actual value of $R$.
Actual values of $R=30 \Omega$
So, measured value of $R$ is
$$ \begin{aligned} & R^{\prime} & =30-(5 % \text { of } 30)=30-\frac{5}{100} \times 30 \\ \Rightarrow \quad & R^{\prime} & =28.5 \Omega \end{aligned} $$
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, $R^{\prime}$ should be equal to parallel combination of $R$ and $R _V$. Mathematically,
$$ \begin{array}{rlrl} R^{\prime} & =\frac{R R _V}{R+R _V}=28.5 \Omega \\ \text { Given, } \quad R & =30 \Omega \Rightarrow \frac{30 R _V}{30+R _V}=28.5 \\ \Rightarrow \quad & 30 R _V & =(28.5 \times 30)+28.5 R _V \\ \Rightarrow \quad 1.5 R _V & =28.5 \times 30 \\ \Rightarrow \quad R _V & =\frac{28.5 \times 30}{1.5}=19 \times 30 \text { or } R _V=570 \Omega \end{array} $$
