The resistance of an ammeter is 13 Omega and | vedclass

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(B) Let $i_{a}$ be the current flowing through the ammeter and $i$ be the total current. So,a current $(i - i_{a})$ will flow through the shunt resist

Vedclass · Accepted Answer

$2 \Omega$

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(B) Let $i_{a}$ be the current flowing through the ammeter and $i$ be the total current. So,a current $(i - i_{a})$ will flow through the shunt resistance.The potential difference across the ammeter and the shunt resistance is the same because they are connected in parallel.Therefore,$i_{a} 	imes R = (i - i_{a}) 	imes S$Rearranging for $S$,we get:$S = \frac{i_{a} R}{i - i_{a}}$Given values are $i_{a} = 100 	ext{ A}$,$i = 750 	ext{ A}$,and $R = 13 \Omega$.Substituting these values into the formula:$S = \frac{100 	imes 13}{750 - 100}$$S = \frac{1300}{650}$$S = 2 \Omega$Thus,the value of the shunt resistance is $2 \Omega$.

The resistance of an ammeter is $13 \Omega$ and its scale is graduated for a current up to $100 \text{ A}$. After an additional shunt has been connected to this ammeter,it becomes possible to measure currents up to $750 \text{ A}$ by this meter. The value of the shunt resistance is

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