Refer to the figure given below. The values of $$I_1$$, $$I_2$$ and $$I_3$$ are _______.

Solution :

Take bottom vertex potential as :

$$V_B = 0$$

Since the right side battery is 5V

Top vertex potential :

$$V_A = 5\text{ V}$$

Let left node potential be $$V_C$$ and middle node potential be $$V_D$$.

Applying KCL at node $$C$$ :

$$\frac{V_C-5}{4} + \frac{V_C}{4} + (V_C-10-V_D) = 0$$

$$V_C - \frac{5}{4} + V_C + 4V_C - 40 - 4V_D = 0$$

$$6V_C - 4V_D = 45 \quad ...(1)$$

Applying KCL at node $$D$$ :

$$\frac{V_D-5}{2} + \frac{V_D}{2} + (V_D -V_C +10)=0$$

$$V_D - \frac{5}{2} + 2V_D - 2V_C +20 =0$$

$$-2V_C +4V_D = -15 \quad ...(2)$$

Adding (1) and (2) :

$$4V_C = 30$$

$$V_C = 7.5\text{ V}$$

Substituting in (2) :

$$-2(7.5)+4V_D = -15$$

$$V_D = 0\text{ V}$$

Current $$I_1$$ :

$$I_1 = \frac{V_D-(V_C-10)}{1}$$

$$= 0-(7.5-10)$$

$$= 2.5\text{ A}$$

Current $$I_3$$ :

$$I_3 = \frac{V_C-0}{4}$$

$$= \frac{7.5}{4}$$

$$= 1.875\text{ A}$$

Current $$I_2$$ :

$$I_2 = \frac{5-7.5}{4} + \frac{5-0}{2}$$

$$= -0.625 + 2.5$$

$$= 1.875\text{ A}$$

Final Answer :

$$I_1 = 2.5\text{ A}$$

$$I_2 = 1.875\text{ A}$$

$$I_3 = 1.875\text{ A}$$