Question:
When a \(15\,\Omega\) resistor is connected in parallel to a galvanometer, its deflection becomes one-fourth. Then the resistance of the galvanometer is
When a \(15\,\Omega\) resistor is connected in parallel to a galvanometer, its deflection becomes one-fourth. Then the resistance of the galvanometer is
Show Hint
For a galvanometer with shunt,
\[
\boxed{
\frac{I_s}{I_g}
=
\frac{R_g}{R_s}.
}
\]
Also,
\[
\boxed{\text{Deflection}\propto I_g.}
\]
A decrease in deflection directly indicates a decrease in the current through the galvanometer.
Updated On: Jul 18, 2026
- \(45\,\Omega\)
- \(60\,\Omega\)
- \(15\,\Omega\)
- \(30\,\Omega\)
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The Correct Option is A
Solution and Explanation
Step 1: Relate galvanometer deflection to current.
The deflection of a galvanometer is directly proportional to the current through it.
Initially,
\[
I_g=I.
\]
After connecting a shunt resistance,
\[
I_g=\frac{I}{4}.
\]
Hence,
\[
\frac{I_s}{I_g}
=
\frac{I-\frac{I}{4}}{\frac{I}{4}}
=
3,
\]
where \(I_s\) is the current through the shunt.
Step 2: Use the current division rule. For parallel branches, \[ \frac{I_s}{I_g} = \frac{R_g}{R_s}. \] Given, \[ R_s=15\,\Omega. \] Therefore, \[ 3=\frac{R_g}{15}, \] \[ R_g=45\,\Omega. \] Hence, \[ \boxed{R_g=45\,\Omega.} \] Therefore, the correct option is \(\boxed{(A)}\).
Step 2: Use the current division rule. For parallel branches, \[ \frac{I_s}{I_g} = \frac{R_g}{R_s}. \] Given, \[ R_s=15\,\Omega. \] Therefore, \[ 3=\frac{R_g}{15}, \] \[ R_g=45\,\Omega. \] Hence, \[ \boxed{R_g=45\,\Omega.} \] Therefore, the correct option is \(\boxed{(A)}\).
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