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

Answer: (C)

20

Answer 2

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.