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 $i_a$ 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, $i_a=100 A , i=750 A , R=13 \Omega$
Hence, $S=\frac{100 \times 13}{750-100}=2 \Omega$