Question:

A galvanometer of resistance 100Ω 100 \, \Omega when connected in series with 400Ω 400 \, \Omega measures a voltage of up to 10V 10 \, V . The value of resistance required to convert the galvanometer into an ammeter to read up to 10A 10 \, A is x×102Ω x \times 10^{-2} \, \Omega . The value of x x is:

Updated On: Mar 22, 2025
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The Correct Option is C

Solution and Explanation

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

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

ig=VRg+Rs=10400+100=10500=20×103A. 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 S in parallel with the galvanometer. The shunt resistance is given by:

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

S=igRgIig=20×103×1001020×103. S = \frac{i_g R_g}{I - i_g} = \frac{20 \times 10^{-3} \times 100}{10 - 20 \times 10^{-3}}.

Simplifying:

S=29.980.2Ω. S = \frac{2}{9.98} \approx 0.2 \, \Omega.

Step 3: Expressing S S in the Required Form
Given that S=x×102Ω S = x \times 10^{-2} \, \Omega , we have:

x=20. x = 20.

Therefore, the value of x x is 20.

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