If {V_{AB}} = 4,V in the given figure,then re | vedclass

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(D) The potential difference between points $A$ and $B$ is given by the formula for parallel branches with cells and resistors: ${V_{AB}} = \frac{\fra

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$20$

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(D) The potential difference between points $A$ and $B$ is given by the formula for parallel branches with cells and resistors: ${V_{AB}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$ Here,$E_1 = 5\,V$,$r_1 = 10\,\Omega$,$E_2 = 2\,V$,and $r_2 = X\,\Omega$. Substituting the values: $4 = \frac{\frac{5}{10} + \frac{2}{X}}{\frac{1}{10} + \frac{1}{X}}$ $4 = \frac{\frac{X + 4}{2X}}{\frac{X + 10}{10X}}$ $4 = \frac{X + 4}{2X} 	imes \frac{10X}{X + 10}$ $4 = \frac{5(X + 4)}{X + 10}$ $4(X + 10) = 5X + 20$ $4X + 40 = 5X + 20$ $X = 20\,\Omega$.

If ${V_{AB}} = 4\,V$ in the given figure,then resistance $X$ will be .............. $\Omega$.

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