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)

$2\, \Omega$

Answer 2

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