A galvanometer of resistance 50 Omega is conn | vedclass

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

Question

(C) Resistance of galvanometer,$R_g = 50 \Omega$. Emf of battery,$V = 3 	ext{ V}$. Resistance connected in series,$R_s = 2950 \Omega$. Total resistan

Vedclass · Accepted Answer

$4450$

Vedclass · Answer

(C) Resistance of galvanometer,$R_g = 50 \Omega$. Emf of battery,$V = 3 	ext{ V}$. Resistance connected in series,$R_s = 2950 \Omega$. Total resistance,$R' = R_g + R_s = 50 + 2950 = 3000 \Omega$. Therefore,the initial current,$I = \frac{V}{R'} = \frac{3}{3000} = 10^{-3} 	ext{ A}$. If the deflection is reduced to $20$ divisions from $30$ divisions,the new current $I'$ will be $I' = I 	imes \frac{20}{30} = 10^{-3} 	imes \frac{2}{3} = \frac{2}{3} 	imes 10^{-3} 	ext{ A}$. Let $R_E$ be the new total resistance of the circuit. Using Ohm's law,$V = I' R_E \Rightarrow R_E = \frac{V}{I'} = \frac{3}{\frac{2}{3} 	imes 10^{-3}} = \frac{9}{2} 	imes 10^3 = 4500 \Omega$. Since $R_E = R_g + R_{new}$,the new series resistance required is $R_{new} = R_E - R_g = 4500 - 50 = 4450 \Omega$.

Similar Questions

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance $G$,its range is $V$. To double its range,a resistance of $1000 \Omega$ is connected in series. The value of $G$ is (in $\Omega$)

An ammeter with a resistance of $2 \,\Omega$ can measure up to $100 \,mA$. What resistance (in $\Omega$) must be connected in series to make it measure up to $5 \,V$?

$A$ galvanometer coil has a resistance of $990 \,\Omega$. It shows a full-scale deflection for a current of $10 \,mA$. Calculate the value of the shunt resistance required to convert it into an ammeter of range $1 \,A$.

Explain the equation of a shunt.

If only $2\%$ of the main current is to be passed through a galvanometer of resistance $G$,then the resistance of the shunt will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module