Question

In the adjacent shown circuit, a voltmeter of internal resistance \(R_v\), when connected across \(B\) and \(C\) reads \(\dfrac{100}{3}\,V\). Neglecting the internal resistance of the battery, the value of \(R_v\) is

• A. \(100\,k\Omega\)
• B. \(75\,k\Omega\)
• C. \(50\,k\Omega\)
• D. \(25\,k\Omega\)

Hint:

When a voltmeter is connected across a resistor, it forms a parallel combination and changes voltage division. Always replace by equivalent resistance.

Answer 1

The Correct Option is C

Step 1: Understand the circuit.
Two resistors of \(50k\Omega\) each are in series across a \(100V\) source.
Point \(B\) is between the two resistors, and voltmeter is connected across \(B\) and \(C\), i.e. across the lower \(50k\Omega\) resistor.
Step 2: Equivalent resistance of lower branch.
Voltmeter resistance \(R_v\) is in parallel with lower \(50k\Omega\):
\[ R_{eq} = \frac{50k \cdot R_v}{50k + R_v} \]
Step 3: Total series resistance.
Upper resistor is \(50k\Omega\), so total:
\[ R_{total} = 50k + R_{eq} \]
Step 4: Use voltage division.
Voltmeter reads potential across lower part:
\[ V_{BC} = \frac{R_{eq}}{50k + R_{eq}}\cdot 100 \]
Given:
\[ V_{BC} = \frac{100}{3} \]
So:
\[ \frac{100}{3} = \frac{R_{eq}}{50k + R_{eq}}\cdot 100 \Rightarrow \frac{1}{3} = \frac{R_{eq}}{50k + R_{eq}} \]
Step 5: Solve for \(R_{eq}\).
\[ 50k + R_{eq} = 3R_{eq} \Rightarrow 50k = 2R_{eq} \Rightarrow R_{eq} = 25k\Omega \]
Step 6: Solve for \(R_v\).
\[ 25k = \frac{50k \cdot R_v}{50k + R_v} \]
Cross multiply:
\[ 25k(50k + R_v) = 50kR_v \]
\[ 1250k^2 + 25kR_v = 50kR_v \]
\[ 1250k^2 = 25kR_v \Rightarrow R_v = 50k\Omega \]
Final Answer:
\[ \boxed{50\,k\Omega} \]