The resistance of an ammeter is 13 Omega and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$2 \Omega$

Vedclass · Answer

(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

Similar Questions

Which of the following pairs of elements cannot form an alloy?

$\left|\begin{array}{lll}24 & 25 & 26 \\ 25 & 26 & 27 \\ 26 & 27 & 27\end{array}\right|$ is equal to

If $|x| < 1$ and $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$,then $x$ is equal to :

The work done in placing a charge of $8 \times 10^{-18} \ C$ on a capacitor of capacitance $100 \ \mu F$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module