Question

A voltmeter of resistance 400 Ω is used to measure the potential difference across the 100 Ω resistor in the circuit shown in the figure (32-E24). (a) What will be the reading of the voltmeter? (b) What was the potential difference across 100 Ω before the voltmeter was connected?

Answer 1

(a) The effective resistance of the circuit,
${\mathrm{R}}_{\mathrm{eff}}=\frac{100\times 400}{500}+200=280\mathrm{\Omega }$
The current through the circuit,
$i=\frac{84}{280}=0.3\mathrm{A}$
Since 100 Ω resistor and 400 Ω resistor are connected in parallel, the potential difference will be same across their ends. Let the current through 100 Ω resistor be i₁ ; then, the current through 400 Ω resistor will be i - i₁.
$$\begin{gathered} 100i_1=400\left(i-i_1\right) \\ \Rightarrow 500i_1=400i \\ \Rightarrow i_i=\frac{4}{5}i=0.24\mathrm{A} \end{gathered}$$
The reading of the voltmeter = 100 × 0.24 = 24 V

(b) Before the voltmeter is connected, the two resistors 100 Ω resistor and 200 Ω resistor are in series.
The effective resistance of the circuit,
$$\begin{gathered} R_{\mathrm{eff}}=\left(200+100\right) \\ =300\mathrm{\Omega } \end{gathered}$$
The current through the circuit,
$i=\frac{84}{300}=0.28\mathrm{A}$
∴ Voltage across the 100 Ω resistor = (0.28 × 100) = 28 V