In the given circuit,the galvanometer reading | vedclass

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(B) The galvanometer reading is zero,which means no current flows through the branch containing the galvanometer and the $2 	ext{ V}$ cell.Therefore,

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$100$

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(B) The galvanometer reading is zero,which means no current flows through the branch containing the galvanometer and the $2 	ext{ V}$ cell.Therefore,the potential difference across the resistor $x$ must be equal to the electromotive force $(EMF)$ of the cell in that branch,which is $2 	ext{ V}$.Let $I$ be the current flowing through the $500 	ext{ } \Omega$ resistor and the resistor $x$.Since the galvanometer branch has zero current,the entire current $I$ flows through the resistor $x$.The potential difference across $x$ is $V_x = I \cdot x = 2 	ext{ V}$.Applying Kirchhoff's Voltage Law $(KVL)$ to the outer loop containing the $12 	ext{ V}$ source,the $500 	ext{ } \Omega$ resistor,and the resistor $x$:$12 - I(500) - I(x) = 0$$12 - 500I - 2 = 0$$10 = 500I$$I = \frac{10}{500} = \frac{1}{50} 	ext{ A}$.Now,substitute the value of $I$ into the equation $I \cdot x = 2$:$\frac{1}{50} \cdot x = 2$$x = 2 \cdot 50 = 100 	ext{ } \Omega$.

In the given circuit,the galvanometer reading is zero. If the internal resistance of the cells is negligible,then the value of $x$ is ............. $\Omega$.

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