Question

The resistance of an ammeter is $13 \,\Omega$ and its scale is graduated for a current up to $100\, A$. 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)

The potential difference across ammeter and shunt is the same.
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.

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