Q62 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
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 _____.
- Correct answerAconnect resistor of $\dfrac{x}{2}\ \Omega$, in series with voltmeter.
- Bconnect resistor of $\dfrac{x}{2}\ \Omega$, in parallel to voltmeter.
- Cconnect a resistor of $x\ \Omega$ in series with voltmeter.
- Dconnect resistor of $2x\ \Omega$ in parallel to voltmeter.
Solution
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.
Hence, the correct answer is option: 'A'.