Question

The voltmeter shown in the figure reads 18 V across the 50 Ω resistor. Find the resistance of the voltmeter.

Answer 1

Let R be the resistance of the voltmeter. For the given circuit, 50 Ω and R are in parallel. The equivalent resistance of these two resistors,

\[R_{eqv} = \frac{50R}{50 + R}\]

This equivalent resistance is now in series with the 24 Ω resistor. The effective resistance of the circuit,

\[R_{eff} = \left( \frac{50R}{50 + R} + 24 \right) \Omega\]

Current through the circuit,

\[i = \frac{30}{R_{eff}}\]

Voltage across the 24 Ω resistor, V₁ = \[i \times 24 = \frac{30}{R_{eff}} \times 24 V\]

Voltage across the 50 Ω resistor, V₂ = \[30 - V_1 = 30 - \frac{30}{R_{eff}} \times 24 V ............\left(\because V_1 + V_2 = 30\right)\]

It is given that V₂ = 18 V

\[\Rightarrow 18 = 30 - \frac{30}{R_{eff}} \times 24\]

\[ \Rightarrow 18 = 30 - \frac{30}{\left( \frac{50R}{50 + R} + 24 \right)} \times 24\]

\[ \Rightarrow \frac{30}{\left( \frac{74R + 1200}{50 + R} \right)} \times 24 = 30 - 18\]

\[ \Rightarrow \frac{30}{\left( \frac{74R + 1200}{50 + R} \right)} = \frac{12}{24} = \frac{1}{2}\]

\[ \Rightarrow 60\left( 50 + R \right) = \left( 74R + 1200 \right)\]

\[ \Rightarrow R \approx 130 \Omega\]