First slide

Moving coil Galvanometer

Question

The resistance of an ammeter is $13 Ω$ and its scale is graduated for a current up to 100 A. After an additional shunt has been connected to this ammeter it becomes possible to measure currents up to 7 50 A by this meter. The value of shunt resistance is

Easy

A

$20 Ω$
B

$2 Ω$
C

$0.2 Ω$
D

$2 k Ω$

Solution

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 the same.
$\Rightarrow i_{a} \times R = (i - i_{a}) \times S$
$or S = \frac{i_{a} R}{i - i_{a}}$ …..(i)
$\begin{array}{l} Given, i_{a} = 100 A, i = 750 A, R = 13 Ω \\ Hence, S = \frac{100 \times 13}{750 - 100} = 2 Ω \end{array}$

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