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,

$$R_{eff}=(\frac{50R}{50+R}+24)\mathrm{\Omega }$$

Current through the circuit,

$$i=\frac{30}{R_{eff}}$$

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

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

It is given that V₂ = 18 V

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

$$\Rightarrow 18=30-\frac{30}{(\frac{50R}{50+R}+24)}\times 24$$

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

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

$$\Rightarrow 60(50+R)=(74R+1200)$$

$$\Rightarrow R\approx 130\mathrm{\Omega }$$