MHT CET · Physics · Current Electricity
An ammeter of resistance ' \(R\) ' gives a full scale deflection when a current of 2 A passes through it. If it is to be converted into an ammeter to measure maximum current of \(10 \mathrm{~A}\), the required shunt is
- A \(\frac{\mathrm{R}}{2}\)
- B \(\frac{\mathrm{R}}{4}\)
- C \(2 \mathrm{R}\)
- D R
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{R}}{4}\)
Step-by-step Solution
Detailed explanation
Consider the diagram below:
Given, \(\mathrm{i}_{\mathrm{g}}=2 \mathrm{~A}\) is current for maximum deflection, \(\mathrm{R}=\mathrm{R}_{\mathrm{g}}\) is the resistance of galvanometer, \(\mathrm{I}=10 \mathrm{~A}\) is the maximum current to be measured by the ammeter.
On considering same potential drop across the galvanometer and the shunt resistance:
\(\Rightarrow\left(I-i_g\right) S=i_g R\).
On plugging in the given values,
\((10 A-2 A) S=(2 A) R\)
The shunt resistance \(S\) can be written as: \(S=\frac{R}{4}\)
Given, \(\mathrm{i}_{\mathrm{g}}=2 \mathrm{~A}\) is current for maximum deflection, \(\mathrm{R}=\mathrm{R}_{\mathrm{g}}\) is the resistance of galvanometer, \(\mathrm{I}=10 \mathrm{~A}\) is the maximum current to be measured by the ammeter.
On considering same potential drop across the galvanometer and the shunt resistance:
\(\Rightarrow\left(I-i_g\right) S=i_g R\).
On plugging in the given values,
\((10 A-2 A) S=(2 A) R\)
The shunt resistance \(S\) can be written as: \(S=\frac{R}{4}\)
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