MHT CET · Physics · Current Electricity
A galvanometer of resistance \(200 \Omega\) is to be converted into an ammeter. The value of shunt resistance which allows \(3 \%\) of the main current through the galvanometer is equal to (nearly)
- A \(7 \Omega\)
- B \(5 \Omega\)
- C \(10 \Omega\)
- D \(6 \Omega\)
Answer & Solution
Correct Answer
(D) \(6 \Omega\)
Step-by-step Solution
Detailed explanation
Given:
Resistance of galvanometer is \(G=200 \Omega\).
Current through the galvanometer is \(i_g=3 \%\) of \(i\).
To convert into an ammeter, the galvanometer with resistance \(G\) has to be shunted with \(S\). The potential drop across the galvanometer and shunt is the same:
\(\begin{aligned} & i_g G=\left(i-i_g\right) S \\ & \Rightarrow \frac{i}{i_g}=1+\frac{G}{S}\end{aligned}\)
Given, \(\frac{i_g}{i}=\frac{3}{100}\)
\(\begin{aligned} & \therefore \frac{100}{3}=1+\frac{200}{S} \\ & \Rightarrow S=\frac{200 \times 3}{97} \approx 6 \Omega\end{aligned}\)
Resistance of galvanometer is \(G=200 \Omega\).
Current through the galvanometer is \(i_g=3 \%\) of \(i\).
To convert into an ammeter, the galvanometer with resistance \(G\) has to be shunted with \(S\). The potential drop across the galvanometer and shunt is the same:
\(\begin{aligned} & i_g G=\left(i-i_g\right) S \\ & \Rightarrow \frac{i}{i_g}=1+\frac{G}{S}\end{aligned}\)
Given, \(\frac{i_g}{i}=\frac{3}{100}\)
\(\begin{aligned} & \therefore \frac{100}{3}=1+\frac{200}{S} \\ & \Rightarrow S=\frac{200 \times 3}{97} \approx 6 \Omega\end{aligned}\)
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