A certain current passing through a galvanome | vedclass

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

Question

(B) Let the initial current be $I$ and the galvanometer resistance be $G$. The deflection is proportional to the current, so $I = k 	imes 100$, where

Vedclass · Accepted Answer

$99$

Vedclass · Answer

(B) Let the initial current be $I$ and the galvanometer resistance be $G$. The deflection is proportional to the current, so $I = k 	imes 100$, where $k$ is a constant.When a shunt $S = 1 \ \Omega$ is connected in parallel, the new current through the galvanometer $I_g$ is given by the current divider rule: $I_g = I \left( \frac{S}{S+G} ight)$.The new deflection is $1$ division, so $I_g = k 	imes 1$.Substituting the values: $k = (k 	imes 100) \left( \frac{1}{1+G} ight)$.Dividing both sides by $k$: $1 = \frac{100}{1+G}$.$1+G = 100$.$G = 99 \ \Omega$.

$A$ certain current passing through a galvanometer produces a deflection of $100$ divisions. When a shunt of $1 \ \Omega$ is connected, the deflection reduces to $1$ division. The galvanometer resistance is: (in $\Omega$)

Similar Questions

The phosphor bronze strip is used in a moving coil galvanometer because

$A$ galvanometer has resistance $G$ and range $V_g$. How much resistance is required to read voltage up to $V$ volt?

In an ammeter,$5 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be:

The net resistance of a voltmeter should be large to ensure that

$A$ voltmeter of resistance $20000\,\Omega$ reads $5 \text{ V}$. To make it read $20 \text{ V}$,the extra resistance required is

Vedclass Test Series

Exam Paper Generator

Online Exam Module