A galvanometer G deflects full scale when a p | vedclass

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

Question

(A) To construct an ammeter from a galvanometer,a shunt resistance $R_S$ is connected in parallel with the galvanometer.The full-scale deflection curr

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(A) To construct an ammeter from a galvanometer,a shunt resistance $R_S$ is connected in parallel with the galvanometer.The full-scale deflection current of the galvanometer $I_g$ is given by $I_g = V / r_g = 0.50 \ V / 25 \ \Omega = 0.02 \ A$.The total current $I$ to be measured is $2.0 \ A$. The current through the shunt resistance $I_S$ is $I_S = I - I_g = 2.0 \ A - 0.02 \ A = 1.98 \ A$.Since the galvanometer and shunt are in parallel,the potential difference across them is the same: $V = I_g \cdot r_g = I_S \cdot R_S$.Therefore,$R_S = (I_g \cdot r_g) / I_S = 0.50 \ V / 1.98 \ A \approx 0.2525 \ \Omega$.The value closest to this is $0.25 \ \Omega$.

Similar Questions

What is a galvanometer? Give its applications.

The resistance of a galvanometer is $25\,\Omega$ and it requires $50\,\mu A$ for full-scale deflection. The value of the shunt resistance required to convert it into an ammeter of $5\,A$ range is:

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 $5 \,V$?

The range of an ammeter of resistance $G$ can be increased from $I$ to $nI$ by connecting

Vedclass Test Series

Exam Paper Generator

Online Exam Module