KCET · Physics · Current Electricity
A certain current on passing through a galvanometer produces a deflection of 100 divisions. When a shunt of one ohm is connected, the deflection reduces to 1 division. The galvanometer resistance is
- A \(100 \Omega\)
- B \(99 \Omega\)
- C \(10 \Omega\)
- D \(9.9 \Omega\)
Answer & Solution
Correct Answer
(B) \(99 \Omega\)
Step-by-step Solution
Detailed explanation
Shunt is connected to the galvanometer
\(\begin{aligned}i_{g} &=\frac{i S}{S+G} \\1 &=\frac{100 \times 1}{(1+G)} \\G &=99 \Omega\end{aligned}\)
\(\Rightarrow \quad G=99 \Omega\)
\(\begin{aligned}i_{g} &=\frac{i S}{S+G} \\1 &=\frac{100 \times 1}{(1+G)} \\G &=99 \Omega\end{aligned}\)
\(\Rightarrow \quad G=99 \Omega\)
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