Q. The resistance of an ammeter is 3$\ \Omega$ 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$\ \Omega$
Explanation
(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,
$ie.,\ {\ i}_{a} \times R = \left( i - i_{a} \right) \times S$
$or\ \ \ S = \frac{i_{a}R}{i - i_{a}}\ \ \ \ \ \ldots\ldots\ldots\ldots(i)$
Given, $i_{a} = 100A,\ \ i = 750A,\ R = 13\Omega$
Hence, $S = \frac{100 \times 13}{750 - 100} = 2\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.