Question

With a resistance '\(X\)' connected in series with a galvanometer of resistance \(100 \Omega\), it acts as a voltmeter of range 0 - 15 V. To double the range, a resistance of \(1500 \Omega\) is to be connected in series with '\(X\)'. The value of '\(X\)' in \(\Omega\) is

• A. \(1000\)
• B. \(1200\)
• C. \(1400\)
• D. \(1600\)

Hint:

The full-scale current of the galvanometer is fixed: \(V=I_g(G+R_{series})\).

Answer 1

The Correct Option is C

Step 1: Use the extra voltage
Doubling the range adds 15 V of extra drop across the additional $1500\ \Omega$.

Step 2: Find the current
$I_g=\dfrac{15}{1500}=0.01$ A. Then $100+X=\dfrac{15}{0.01}=1500$, so $X=1400\ \Omega$, option (C).

Final Answer:
$X=1400\ \Omega$, option (C). \[ \boxed{1400\ \Omega} \]