In the given circuit diagram,the galvanometer | vedclass

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(A) Given that the galvanometer reading is zero,$I_g = 0$.This implies that the current $I$ flowing through the $400 \ \Omega$ resistor and the resist

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$100$

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(A) Given that the galvanometer reading is zero,$I_g = 0$.This implies that the current $I$ flowing through the $400 \ \Omega$ resistor and the resistor $X$ is the same.The total resistance of the circuit loop is $R_{total} = 400 + X$.Using Ohm's law,the current $I$ is given by $I = \frac{10}{400 + X}$.The potential difference across the resistor $X$ is equal to the $EMF$ of the second cell,which is $2 \ V$.Therefore,$I \cdot X = 2$.Substituting the expression for $I$: $\frac{10 \cdot X}{400 + X} = 2$.$10X = 2(400 + X)$.$10X = 800 + 2X$.$8X = 800$.$X = 100 \ \Omega$.

In the given circuit diagram,the galvanometer reading is zero. If the internal resistance of the cell is neglected,then the value of $X$ is ......... $\Omega$.

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