JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2026
- Medium
- MCQ
A voltmeter with internal resistance of \(x\ \Omega\) can be used to measure upto \(20\) V. In order to increase its measuring range to \(30\) V, the required modification is to _____.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) connect resistor of \(\dfrac{x}{2}\ \Omega\), in series with voltmeter.
Step-by-step Solution
Detailed explanation
Let the maximum current for full-scale deflection of the voltmeter be \(I_g\).
\(I_g = \dfrac{V_1}{R_v} = \dfrac{20}{x}\)
To increase the measuring range to \(V_2 = 30\) V, a resistance \(R\) must be connected in series with the voltmeter.
\(V_2 = I_g(R_v + R)\)
\(30 = \dfrac{20}{x}(x + R)\)
\(30x = 20x + 20R\)
\(10x = 20R\)
\(R = \dfrac{x}{2}\ \Omega\)
Thus, a resistor of \(\dfrac{x}{2}\ \Omega\) should be connected in series with the voltmeter.
Answer: connect resistor of \(\dfrac{x}{2}\ \Omega\), in series with voltmeter.
\(I_g = \dfrac{V_1}{R_v} = \dfrac{20}{x}\)
To increase the measuring range to \(V_2 = 30\) V, a resistance \(R\) must be connected in series with the voltmeter.
\(V_2 = I_g(R_v + R)\)
\(30 = \dfrac{20}{x}(x + R)\)
\(30x = 20x + 20R\)
\(10x = 20R\)
\(R = \dfrac{x}{2}\ \Omega\)
Thus, a resistor of \(\dfrac{x}{2}\ \Omega\) should be connected in series with the voltmeter.
Answer: connect resistor of \(\dfrac{x}{2}\ \Omega\), in series with voltmeter.
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