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, $S=\frac{100 \times 13}{750-100}$
$=2 \Omega$