Question:
A galvanometer having a coil resistance of \(100\ \Omega\) gives a full scale deflection when a current of \(1\) mA is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving a full scale deflection of \(10\) A, is nearly
A galvanometer having a coil resistance of \(100\ \Omega\) gives a full scale deflection when a current of \(1\) mA is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving a full scale deflection of \(10\) A, is nearly
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A shunt in parallel with the galvanometer shares the same potential difference and carries the extra current.
Updated On: Nov 12, 2026
- \(3\ \Omega\)
- \(2\ \Omega\)
- \(0.1\ \Omega\)
- \(0.01\ \Omega\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
A galvanometer becomes an ammeter when a low resistance (shunt) \(S\) is connected in parallel with it.
Step 2: Parallel condition
Same potential difference: \(I_gG=(I-I_g)S\), so \(S=\dfrac{I_gG}{I-I_g}\).
Step 3: Calculation
\(S=\dfrac{10^{-3}\times 100}{10-10^{-3}}=\dfrac{0.1}{9.999}\approx 0.01\ \Omega\)
Step 4: Result
The shunt is about \(0.01\ \Omega\), option (D).
Final Answer:
A shunt of about \(0.01\ \Omega\) is needed. \[ \boxed{\text{(D) }0.01\ \Omega} \]
A galvanometer becomes an ammeter when a low resistance (shunt) \(S\) is connected in parallel with it.
Step 2: Parallel condition
Same potential difference: \(I_gG=(I-I_g)S\), so \(S=\dfrac{I_gG}{I-I_g}\).
Step 3: Calculation
\(S=\dfrac{10^{-3}\times 100}{10-10^{-3}}=\dfrac{0.1}{9.999}\approx 0.01\ \Omega\)
Step 4: Result
The shunt is about \(0.01\ \Omega\), option (D).
Final Answer:
A shunt of about \(0.01\ \Omega\) is needed. \[ \boxed{\text{(D) }0.01\ \Omega} \]
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