MHT CET · Physics · Current Electricity
When a resistance of \(100 \Omega\) is connected in series with a galvanometer of resistance \(G\), its range is V . To double its range a resistance of \(1000 \Omega\) is connected in series. The value of \(G\) is
- A \(900 \Omega\)
- B \(300 \Omega\)
- C \(200 \Omega\)
- D \(100 \Omega\)
Answer & Solution
Correct Answer
(A) \(900 \Omega\)
Step-by-step Solution
Detailed explanation
When a resistance of \(100 \Omega\) is connected in series
current, \(i=\frac{2 V}{100+R}\) \(\qquad\) when a resistance of \(1000 \Omega\) is connected in series the its range double
current, \(\quad i=\frac{2 V}{1100+R}\)
From above equations
\(\begin{aligned}
& \frac{V}{100+R}=\frac{2 V}{1100+R} \\
& R=900 \Omega
\end{aligned}\)
current, \(i=\frac{2 V}{100+R}\) \(\qquad\) when a resistance of \(1000 \Omega\) is connected in series the its range double
current, \(\quad i=\frac{2 V}{1100+R}\)
From above equations
\(\begin{aligned}
& \frac{V}{100+R}=\frac{2 V}{1100+R} \\
& R=900 \Omega
\end{aligned}\)
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