Question:
A galvanometer of resistance G is shunted with a resistance of 10% of G. The part of the total current that flows through the galvanometer is
A galvanometer of resistance G is shunted with a resistance of 10% of G. The part of the total current that flows through the galvanometer is
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Logic Tip: A shunt allows a larger current to bypass the delicate galvanometer, effectively converting it into an ammeter.
Updated On: Apr 28, 2026
- $\frac{1}{11}I$
- $\frac{2}{11}I$
- $\frac{1}{10}I$
- $\frac{1}{5}I$
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The Correct Option is A
Solution and Explanation
Concept:
In a shunted galvanometer, current is divided between the galvanometer and the shunt resistance in parallel.
Step 1: Identify the shunt resistance.
Shunt resistance $S = 10%$ of $G = 0.1G$.
Step 2: Use the current division formula.
The current through the galvanometer $I_{g}$ is: $$\frac{I_{g{I}=\frac{S}{S+G}$$
Step 3: Substitute the values.
$$\frac{I_{g{I}=\frac{0.1G}{0.1G+G}=\frac{0.1G}{1.1G}=\frac{1}{11}$$ $$I_{g}=\frac{1}{11}I$$
In a shunted galvanometer, current is divided between the galvanometer and the shunt resistance in parallel.
Step 1: Identify the shunt resistance.
Shunt resistance $S = 10%$ of $G = 0.1G$.
Step 2: Use the current division formula.
The current through the galvanometer $I_{g}$ is: $$\frac{I_{g{I}=\frac{S}{S+G}$$
Step 3: Substitute the values.
$$\frac{I_{g{I}=\frac{0.1G}{0.1G+G}=\frac{0.1G}{1.1G}=\frac{1}{11}$$ $$I_{g}=\frac{1}{11}I$$
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