In the given circuit,the value of left|frac{I | vedclass

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(D) Let the potential at the node between the $10\,V$ and $20\,V$ batteries be $V_x = 20\,V$. The potential at the left junction is $10\,V$ and the po

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$2$

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(D) Let the potential at the node between the $10\,V$ and $20\,V$ batteries be $V_x = 20\,V$. The potential at the left junction is $10\,V$ and the potential at the right end of the bottom resistor is $0\,V$.For the two parallel $10\,\Omega$ resistors,the potential difference across each is $V_x - 10\,V = 20\,V - 10\,V = 10\,V$.Thus,$I_1 = \frac{10\,V}{10\,\Omega} = 1\,A$ and $I_2 = \frac{10\,V}{10\,\Omega} = 1\,A$.For the bottom branch,the potential difference across the $10\,\Omega$ resistor is $10\,V - 0\,V = 10\,V$.Thus,$I_3 = \frac{10\,V}{10\,\Omega} = 1\,A$.Now,calculating the required value:$\left|\frac{I_1+I_3}{I_2}\right| = \left|\frac{1\,A + 1\,A}{1\,A}\right| = \left|\frac{2}{1}\right| = 2$.

In the given circuit,the value of $\left|\frac{I_1+I_3}{I_2}\right|$ is

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