Using Kirchhoff's laws of current and voltage, find the current in $2 \; \Omega$ resistor in the given circuit.

Given,

Applying Kirchhoff's current law at point $'B'$.

$I_3 = I_1  + I_2$ .......... (i)

Now, Applying Kirchhoff's voltage law in close loop ABEFA

$35 = I_1 * 3 + I_3 * 2$

$35 = 3 \; I_1 + 2 * (I_1 + I_2)$

or, $3\;I_1 + 2 \; I_2 + 2 \; I_2 = 35$

or, $5\; I_1 + 2 \; I_2 = 35$ ......... (ii)

Again, Applying Kirchhoff's Voltage law in close loop CBEDC,

$40 = I_2 * 4 + I_3 * 2$

or, $40 = I_2 * 4 + I_3 * 2$

or, $40 = 4\; I_2 + 2 (I_1 + I_2)$

or, $2\; I_1 + 6 \; I_2 = 40$ .......... (iii)

Solving equation (ii) and (iii) we get,

$I_1 = 5\;A$       and       $I_2 = 5\; A$

So current through $2 \;\Omega = I_3 = I_1 + I_2 = 5 + 5 = 10 \; A$

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