In the adjoining circuit,the battery E_1 has | vedclass

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(B) Given that the galvanometer $G$ reads zero,no current flows through the branch containing $E_2$ and the galvanometer.Therefore,the current $I$ flo

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$100$

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(B) Given that the galvanometer $G$ reads zero,no current flows through the branch containing $E_2$ and the galvanometer.Therefore,the current $I$ flows only through the loop containing $E_1$,the $500 \, \Omega$ resistor,and the resistor $X$.Using Ohm's law for the circuit:$I = \frac{E_1}{500 + X}$The potential difference across the resistor $X$ must be equal to the emf of the battery $E_2$ for the galvanometer to read zero:$V_X = I \cdot X = E_2$Substituting the value of $I$:$\left( \frac{E_1}{500 + X} \right) \cdot X = E_2$Given $E_1 = 12 \, V$ and $E_2 = 2 \, V$:$\frac{12 \cdot X}{500 + X} = 2$$12X = 2(500 + X)$$12X = 1000 + 2X$$10X = 1000$$X = 100 \, \Omega$

In the adjoining circuit,the battery $E_1$ has an emf of $12 \, V$ and zero internal resistance,while the battery $E_2$ has an emf of $2 \, V$. If the galvanometer $G$ reads zero,then the value of the resistance $X$ (in $\Omega$) is

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