Refer to the figure given below. The values of $I_1, I_2$ and $I_3$ are ______.
Answer: (A) $I_1 = 2.5$ A, $I_2 = 1.875$ A, $I_3 = 1.875$ A
Take the bottom node as $0$ V. The $5$ V cell makes the top node $5$ V. Let the left node be $V_L$ and the middle node $V_M$.
The middle branch ($10$ V cell, positive towards the left, with $1\ \Omega$) carries $I_1$ from the middle node to the left node: $V_L = V_M + 10 - I_1$.
KCL at the left node ($I_1$ in, out through the two $4\ \Omega$): $I_1 = \dfrac{V_L - 5}{4} + \dfrac{V_L}{4}$.
KCL at the middle node: $\dfrac{V_M - 5}{2} + \dfrac{V_M}{2} + I_1 = 0$.
Solving: $V_M = 0$, $V_L = 7.5$ V, $I_1 = 2.5$ A.
$I_3 = \dfrac{7.5}{4} = 1.875$ A. At the top node, $0.625$ A arrives from the left and $\dfrac{5}{2} = 2.5$ A leaves through the upper $2\ \Omega$, so $I_2 = 2.5 - 0.625 = 1.875$ A.
Solution by Sreeraj P, M.Sc Physics