MHT CET · Physics · Current Electricity
A galvanometer of resistance \(50 \Omega\) is connected to a battery of \(3 \mathrm{~V}\) along with a resistance of \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
- A \(5050 \Omega\)
- B \(5550 \Omega\)
- C \(6050 \Omega\)
- D \(4450 \Omega\)
Answer & Solution
Correct Answer
(D) \(4450 \Omega\)
Step-by-step Solution
Detailed explanation
Current through the galvanometer
\(
\begin{aligned}
I=\frac{3}{(50+2950)} &=10^{-3} \mathrm{~A} \\
\text { Current for } 30 \text { divisions } &=10^{-3} \mathrm{~A} \\
\text { Current for } 20 \text { divisions } &=\frac{10^{-3}}{30} \times 20 \\
&=\frac{2}{3} \times 10^{-3} \mathrm{~A}
\end{aligned}
\)
For the same deflection to obtain for 20 divisions, let resistance added be \(R\)
\(
\begin{array}{lc}
\therefore \quad \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 R)} \\
\text { or } \quad R=4450 \Omega
\end{array}
\)
\(
\begin{aligned}
I=\frac{3}{(50+2950)} &=10^{-3} \mathrm{~A} \\
\text { Current for } 30 \text { divisions } &=10^{-3} \mathrm{~A} \\
\text { Current for } 20 \text { divisions } &=\frac{10^{-3}}{30} \times 20 \\
&=\frac{2}{3} \times 10^{-3} \mathrm{~A}
\end{aligned}
\)
For the same deflection to obtain for 20 divisions, let resistance added be \(R\)
\(
\begin{array}{lc}
\therefore \quad \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 R)} \\
\text { or } \quad R=4450 \Omega
\end{array}
\)
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