The resistance of an ammeter is 13 Ω 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

20 Ω

2 Ω

0.2 Ω

2 KΩ

Solution

2 Ω

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

i.e i_a × R = ( i - i_a ) × S

$\Rightarrow S = \frac{i_{a} R}{i - i_{a}}$

Given:- i_a = 100 A, i = 750 A, R = 13 Ω

Hence

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

S = 2Ω

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