MHT CET · Physics · Current Electricity
The resistance of an ammeter is \(13 \Omega\) and its scale is graduated for a current upto \(100 \mathrm{~A}\). After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto \(750 \mathrm{~A}\) by this meter. The value of shunt resistance is
- A \(20 \Omega\)
- B \(2 \Omega\)
- C \(0.2 \Omega\)
- D \(2 \mathrm{k} \Omega\)
Answer & Solution
Correct Answer
(B) \(2 \Omega\)
Step-by-step Solution
Detailed explanation
Let \(i_{a}\) is the current flowing through ammeter and \(i\) is the total current. So, a current \(i-i_{a}\) will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same.

ie, \(\quad i_{a} \times R=\left(i-i_{a}\right) \times S\)
Or
\(S=\frac{i_{a} R}{i-i_{a}} ...(i)\)
Given, \(i_{a}=100 \mathrm{~A}, i=750 \mathrm{~A}, R=13 \Omega\)
Hence, \(\quad S=\frac{100 \times 13}{750-100}=2 \Omega\)
Potential difference across ammeter and shunt resistance is same.

ie, \(\quad i_{a} \times R=\left(i-i_{a}\right) \times S\)
Or
\(S=\frac{i_{a} R}{i-i_{a}} ...(i)\)
Given, \(i_{a}=100 \mathrm{~A}, i=750 \mathrm{~A}, R=13 \Omega\)
Hence, \(\quad S=\frac{100 \times 13}{750-100}=2 \Omega\)
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