Question

The resistance of an ammeter is $13\, \Omega$ and its scale is graduated for a current upto $100\, A$. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto $750\, A$ by this meter. The value of shunt resistance is

• A. $20\, \Omega $
• B. $2\, \Omega $
• C. $0.2\,\Omega $
• D. $2\,k\Omega $

Answer 1

Answer: (B)

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, $i_{a} \times R=\left(i-i_{a}\right) \times S$
or $S=\frac{i_{a} R}{i-i_{a}}$ ...(i)
Given, $i_{a}=100\, A , i=750\, A,\, R=13\, \Omega$
Hence, $S=\frac{100 \times 13}{750-100}=2\, \Omega$