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

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1785820693/questions/llavfxy0ubdlkgcpjj17.png)

Question

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

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1785820693/questions/llavfxy0ubdlkgcpjj17.png)

Accepted Answer

A. 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 $I_3$:

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

10. Using this in equation (v):

$$I_2 = 1.875	ext{ A}$$

11. Using this in equation (vi):

$$I_1 = 2.5	ext{ A}$$

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

Refer to the figure given below. The values of I1,I2I_1, I_2I1​,I2​ and I3I_3I3​ are ____ .

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