In the adjoining circuit,the battery E_1 has | vedclass

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

Question

(B) For the galvanometer $G$ to read zero,no current flows through the branch containing $G$ and $E$.This implies that the potential difference across

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) For the galvanometer $G$ to read zero,no current flows through the branch containing $G$ and $E$.This implies that the potential difference across the resistance $X$ must be equal to the $e.m.f.$ of the battery $E$.Let $I$ be the current flowing through the loop containing $E_1$,$500 \, \Omega$,and $X$.Using Ohm's law,the current $I$ is given by $I = \frac{E_1}{500 + X}$.The potential difference across $X$ is $V_X = I \cdot X = \left( \frac{E_1}{500 + X} \right) X$.Since $V_X = E$,we have $\left( \frac{12}{500 + X} \right) X = 2$.Dividing both sides by $2$,we get $\left( \frac{6}{500 + X} \right) X = 1$.$6X = 500 + X$.$5X = 500$.$X = 100 \, \Omega$.

In the adjoining circuit,the battery $E_1$ has an $e.m.f.$ of $12 \, V$ and zero internal resistance,while the battery $E$ has an $e.m.f.$ of $2 \, V$. If the galvanometer $G$ reads zero,then the value of the resistance $X$ in $\Omega$ is

Similar Questions

Consider the circuit shown. If all the cells have negligible internal resistance,what will be the current through the $2 \Omega$ resistor when steady state is reached (in $A$)?

The value of $I$ in the figure shown below is: (in $\text{ A}$)

When the switch $S$, in the circuit shown, is closed, then the value of current $i$ will be ................. $A$.

In the circuit shown in the figure,$P \neq R$. The reading of the galvanometer remains the same with switch $S$ open or closed. Then

In the circuit shown,the current in the $1\,\Omega$ resistor is

Vedclass Test Series

Exam Paper Generator

Online Exam Module