Question

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

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

Answer 1

Answer: (B)

Let $i_a$ be the current flowing through ammeter and $i$ the total current. So, a current $i-i_a$ will flow through shunt resistance.

Potential difference across ammeter and shunt resistance is same,
i.e., $i_a\times R=\left(i-i_a\right)\times S$
or $S=\frac{i_{a}R}{i-i_a}.....$(i)
Given, $i_a=100A,i=750A,R=13\Omega$
Hence, $S=\frac{100\times 13}{750-100}=2\Omega$