Question

A galvanometer of resistance \( 100 \, \Omega \) when connected in series with \( 400 \, \Omega \) measures a voltage of up to \( 10 \, V \). The value of resistance required to convert the galvanometer into an ammeter to read up to \( 10 \, A \) is \( x \times 10^{-2} \, \Omega \). The value of \( x \) is:

• A. 2
• B. 800
• C. 20
• D. 200

Answer 1

The Correct Option is C

Given:
- Resistance of galvanometer: \( R_g = 100 \, \Omega \)
- Series resistance: \( R_s = 400 \, \Omega \)
- Voltage measured: \( V = 10 \, \text{V} \)

Step 1: Calculating the Current through the Galvanometer
The current through the galvanometer, \( i_g \), is given by:

\[ i_g = \frac{V}{R_g + R_s} = \frac{10}{400 + 100} = \frac{10}{500} = 20 \times 10^{-3} \, \text{A}. \]

Step 2: Converting the Galvanometer to an Ammeter
To convert the galvanometer into an ammeter that can read up to 10 A, we need to connect a shunt resistance \( S \) in parallel with the galvanometer. The shunt resistance is given by:

\[ i_g R_g = (I - i_g) S, \]
where \( I = 10 \, \text{A} \) is the total current.
Rearranging for \( S \):

\[ S = \frac{i_g R_g}{I - i_g} = \frac{20 \times 10^{-3} \times 100}{10 - 20 \times 10^{-3}}. \]

Simplifying:

\[ S = \frac{2}{9.98} \approx 0.2 \, \Omega. \]

Step 3: Expressing \( S \) in the Required Form
Given that \( S = x \times 10^{-2} \, \Omega \), we have:

\[ x = 20. \]

Therefore, the value of \( x \) is 20.