MHT CET · Physics · Current Electricity
When a resistance of \(100 \Omega\) is connected in series with a galvanometer of resistance \(R\), its range is
V. To double its range, a resistance of \(1000 \Omega\) is connected in series. Find \(R\).
- A \(700 \Omega\)
- B \(800 \Omega\)
- C \(900 \Omega\)
- D \(100 \Omega\)
Answer & Solution
Correct Answer
(C) \(900 \Omega\)
Step-by-step Solution
Detailed explanation
When a resistance of \(100 \Omega\) is connected in series
\(
\text { current, } i=\frac{V}{100+R}
\)
when a resistance of \(1000 \Omega\) is connected in
series, the its range double
\(
\text { current, } i=\frac{2 V}{1100+R}
\)
From Eqs. (i) and (ii)
\(
\begin{aligned}
\frac{V}{100+R} &=\frac{2 V}{1100+R} \\
R &=900 \Omega
\end{aligned}
\)
\(
\text { current, } i=\frac{V}{100+R}
\)
when a resistance of \(1000 \Omega\) is connected in
series, the its range double
\(
\text { current, } i=\frac{2 V}{1100+R}
\)
From Eqs. (i) and (ii)
\(
\begin{aligned}
\frac{V}{100+R} &=\frac{2 V}{1100+R} \\
R &=900 \Omega
\end{aligned}
\)
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