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:
- 2
- 800
- 20
- 200
The Correct Option is C
Approach Solution - 1
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.
Approach Solution -2
Resistance of galvanometer, G = 100 Ω
Series resistance, R = 400 Ω
Voltage range, V = 10 V
Required current range for ammeter, I = 10 A
Step 2: Find the current needed for full-scale deflection of galvanometer.
When the galvanometer is connected in series with 400 Ω and reads 10 V:
Ig = V / (R + G) = 10 / (400 + 100) = 10 / 500 = 0.02 A
Step 3: Find shunt resistance (S) required to convert it into an ammeter.
Let S be the resistance connected in parallel with the galvanometer.
The total current through the ammeter is 10 A, while only 0.02 A passes through the galvanometer.
Hence, current through shunt = I - Ig = 10 - 0.02 = 9.98 A
Since both are parallel, voltage across galvanometer = voltage across shunt:
Ig G = (I - Ig) S
Substitute values:
0.02 × 100 = 9.98 × S
S = (2) / (9.98) ≈ 0.2004 Ω
Step 4: Express S in the given form.
S = 0.2004 Ω = 20.04 × 10⁻² Ω
Hence, x = 20
Final Answer: 20
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