Question

A cell of emf 90 V is connected across a series combination of two resistors, each of 100 \( \Omega \) resistance. A voltmeter of resistance 400 \( \Omega \) is used to measure the potential difference across each resistor. The reading of the voltmeter will be:

• A. \(45 \, V\)
• B. \(40 \, V\)
• C. \(80 \, V\)
• D. \(90\, V\)

Answer 1

The Correct Option is B

Resistance of each resistor, \( R = 100 \, \Omega \),   Resistance of voltmeter, \( R_v = 400 \, \Omega \)

EMF of the cell, \( \text{EMF} = 90 \, \text{V} \)

The equivalent resistance across one resistor and the voltmeter is:

\[ R_{\text{eq}} = \frac{R_v \cdot R}{R_v + R} = \frac{400 \cdot 100}{400 + 100} = \frac{40000}{500} = 80 \, \Omega \]

The total resistance in the circuit is:

\[ R_{\text{total}} = R + R_{\text{eq}} = 100 + 80 = 180 \, \Omega \]

The current through the circuit is:

\[ I = \frac{\text{EMF}}{R_{\text{total}}} = \frac{90}{180} = 0.5 \, \text{A} \]

The potential drop across the resistor-voltmeter combination is:

\[ V = I \times R_{\text{eq}} = 0.5 \times 80 = 40 \, \text{V} \]

Thus, the reading of the voltmeter is:

\[ \boxed{40 \, \text{V}} \]

Answer 2

The correct answer is (B) : 40 V

$R_{\text{eq}}=\frac{400\times 100}{500}+100$
$=180\Omega$
$i=\frac{90}{180}=\frac{1}{2}A$
Reading $=\frac{1}{2}\times \frac{400}{500}\times 100$
$=40\text{volt}$