The value of shunt resistance that allows only 10% of the main current through the galvanometer of resistance \( 99 \Omega \) is:
The value of shunt resistance that allows only 10% of the main current through the galvanometer of resistance \( 99 \Omega \) is:
Show Hint
- \( 9 \Omega \)
- \( 4 \Omega \)
- \( 2 \Omega \)
- \( 11 \Omega \)
The Correct Option is D
Approach Solution - 1
To solve for the shunt resistance \( S \) that allows only 10% of the main current \( I \) through the galvanometer of resistance \( R_g = 99 \, \Omega \), we use the formula for a galvanometer shunt:
\( I_g = \frac{I}{10} \) where \( I_g \) is the current through the galvanometer.
The total current through the circuit divides between the galvanometer and the shunt, hence:
\( I_g R_g = (I - I_g)S \)
Substituting \( I_g = \frac{I}{10} \):
\( \frac{I}{10} \cdot 99 = \left(I - \frac{I}{10}\right)S \)
Simplifying, we get:
\( \frac{99I}{10} = \frac{9I}{10}S \)
Divide both sides by \( \frac{9I}{10} \):
\( S = \frac{99}{9} = 11 \, \Omega \)
Thus, the shunt resistance that allows only 10% of the current through the galvanometer is \( 11 \Omega \).
Approach Solution -2
Step 1: Understanding Shunt Resistance A shunt resistance \( S \) is connected in parallel with a galvanometer to allow a fraction of the total current to pass through it, protecting the galvanometer from excessive current. The relation for the shunt resistance is given by: \[ S = \frac{G I_g}{I - I_g} \] where: - \( G = 99 \Omega \) (Galvanometer resistance), - \( I_g = 0.1 I \) (10\% of the main current passes through the galvanometer), - \( I - I_g = 0.9 I \) (90\% of the current passes through the shunt).
Step 2: Calculating the Shunt Resistance Using the formula: \[ S = \frac{99 \times 0.1 I}{0.9 I} \] \[ S = \frac{99 \times 0.1}{0.9} \] \[ S = \frac{9.9}{0.9} = 11 \Omega \]
Step 3: Conclusion Thus, the correct value of the shunt resistance is \( 11 \Omega \).
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