Current Electricity — JEE Main practice

1. Kirchhoff's Current Law (KCL) says the current entering a junction equals the current leaving it. Applying this at junction A, where $I_1$ splits between $I_3$ and the branch toward D ($I_{AD}$):

$I_{AD} = I_1 - I_3$

2. Applying KCL again at junction D, where $I_{AD}$ combines with $I_2$ to give $I_{DB}$:

$I_{DB} = I_1 - I_3 + I_2$

3. Using Kirchhoff's Voltage Law (KVL), which says the sum of potential drops around any closed loop is zero, in loop I:

$-7I_1 - 2I_2 + 6I_3 = -10 \quad ...(i)$

4. Applying KVL in loop II:

$2I_2 - 6I_3 - I_1 = -10 \quad ...(ii)$

5. Applying KVL in loop III:

$2I_1 + 4I_2 - 4I_3 = 5 \quad ...(iii)$

6. Solving equation (iii) for $I_1$:

$I_1 = \frac{5 - 4I_2 + 4I_3}{2} \quad ...(iv)$

7. Substituting (iv) into (ii) and simplifying gives a link between $I_2$ and $I_3$:

$I_2 = 2I_3 - \frac{15}{8} \quad ...(v)$

8. Substituting (v) back into (iv) expresses $I_1$ purely in terms of $I_3$:

$I_1 = \frac{25}{4} - 2I_3 \quad ...(vi)$

9. Substituting (v) and (vi) into equation (i) and solving for $I_3$:

$16I_3 = 30 \Rightarrow I_3 = 1.875\text{ A}$

10. Using this in equation (v):

$I_2 = 1.875\text{ A}$

11. Using this in equation (vi):

$I_1 = 2.5\text{ A}$

Hence, $I_1 = 2.5$ A, $I_2 = 1.875$ A, $I_3 = 1.875$ A, so the answer is option A.