A galvanometer of resistance 50 Omega is con | vedclass

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

Question

(D) The current $I$ flowing through the galvanometer is proportional to the deflection $\phi$ in divisions,i.e.,$I \propto \phi$.Using Ohm's law,$I =

Vedclass · Accepted Answer

$7950$

Vedclass · Answer

(D) The current $I$ flowing through the galvanometer is proportional to the deflection $\phi$ in divisions,i.e.,$I \propto \phi$.Using Ohm's law,$I = \frac{V}{R + G}$,where $V = 8 \ V$,$G = 50 \ \Omega$,and $R$ is the series resistance.For the first case: $I_1 = \frac{V}{R_1 + G}$ with $\phi_1 = 30$ divisions and $R_1 = 3950 \ \Omega$.For the second case: $I_2 = \frac{V}{R_2 + G}$ with $\phi_2 = 15$ divisions.Taking the ratio: $\frac{I_1}{I_2} = \frac{\phi_1}{\phi_2} \implies \frac{R_2 + G}{R_1 + G} = \frac{30}{15} = 2$.Substituting the values: $\frac{R_2 + 50}{3950 + 50} = 2$.$\frac{R_2 + 50}{4000} = 2$.$R_2 + 50 = 8000$.$R_2 = 7950 \ \Omega$.

Similar Questions

Give the unit of shunt.

$A$ galvanometer of resistance $G$ is shunted by a resistance $S$. To keep the main current in the circuit unchanged,the resistance to be put in series with the galvanometer is

$A$ galvanometer has a resistance of $80 \Omega$ and it is shunted with a resistance of $20 \Omega$. If $20 \%$ of the main current flows through the galvanometer, what is the value of the main current (in $\text{ A}$)?

$A$ galvanometer of resistance $40 \Omega$ gives a deflection of $10 \text{ divisions per } mA$. There are $50 \text{ divisions}$ on the scale. The maximum current that can pass through the circuit when a shunt resistance of $2 \Omega$ is connected is: (in $\text{ mA}$)

To reduce the current flowing through a coil of resistance $90\, \Omega$ by $90\%$,a resistor of what value in $\Omega$ must be connected in parallel?

Vedclass Test Series

Exam Paper Generator

Online Exam Module