Question

When a resistance of 100 $\Omega$ is connected in series with a galvanometer of resistance 'G', its range is 'V'. To double its range, a resistance of 1000 $\Omega$ is connected in series. The value of 'G' is ______.

• A. 400 $\Omega$
• B. 800 $\Omega$
• C. 1000 $\Omega$
• D. 1200 $\Omega$

Hint:

When an instrument's range is extended $n$ times ($V_{new} = n \cdot V_{old}$), the new required total series resistance must simply be $n$ times the old total series resistance! $G + R_2 = n(G + R_1)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A galvanometer is converted into a voltmeter by placing a high resistance in series with it. We are given two distinct calibration scenarios with different series resistors and resulting voltage ranges. We must calculate the internal resistance of the galvanometer ($G$).

Step 2: Detailed Explanation:
The formula for a voltmeter's total voltage range ($V$) is defined by Ohm's law. The maximum voltage it can measure occurs when the galvanometer reaches its full-scale deflection current ($I_g$).
$V = I_g \times R_{\text{total}}$
Since the galvanometer ($G$) and the multiplier resistor ($R$) are in series, $R_{\text{total}} = G + R$.
$V = I_g (G + R)$
Scenario 1:
A series resistance of $100 \ \Omega$ gives a range of $V$.
$V = I_g (G + 100)$ --- (Equation 1)
Scenario 2:
A series resistance of $1000 \ \Omega$ doubles the range to $2V$.
$2V = I_g (G + 1000)$ --- (Equation 2)
We have a system of two equations. The easiest way to solve for $G$ is to divide Equation 2 by Equation 1. This cancels out the unknown full-scale deflection current $I_g$ completely!
$\frac{2V}{V} = \frac{I_g (G + 1000)}{I_g (G + 100)}$
$2 = \frac{G + 1000}{G + 100}$
Cross-multiply to solve the linear equation:
$2 (G + 100) = G + 1000$
$2G + 200 = G + 1000$
Bring all $G$ terms to the left and constants to the right:
$2G - G = 1000 - 200$
$G = 800 \ \Omega$

Step 3: Final Answer:
The value of 'G' is 800 $\Omega$, matching option (b).