Increasing Range of Voltmeter with Internal Resistance

To increase the voltage range of a voltmeter, a resistor (known as a multiplier) must be connected in series with the original voltmeter.

1. Initial State:

• Internal resistance of the voltmeter, $R_g = x\ \Omega$
• Initial maximum measurable voltage, $V = 20\text{ V}$

The maximum current $I_g$ passing through the voltmeter for full-scale deflection is given by Ohm's Law: $I_g = \frac{V}{R_g} = \frac{20}{x}\text{ A}$

2. Modified State:

• Desired maximum voltage range, $V' = 30\text{ V}$
• Let the required additional series resistance be $R$.

The total resistance of the modified voltmeter is: $R_{\text{total}} = R_g + R = x + R$

Since the maximum full-scale current $I_g$ must remain the same: $V' = I_g \cdot R_{\text{total}}$

Substituting the values of $V'$ and $I_g$: $30 = \left(\frac{20}{x}\right) \cdot (x + R)$

Solving for $R$: $30x = 20(x + R)$ $30x = 20x + 20R$ $10x = 20R$ $R = \frac{x}{2}\ \Omega$

Therefore, the required modification is to connect a resistor of $\frac{x}{2}\ \Omega$ in series with the voltmeter.

Correct Option: A