Question

Find current $i_2$ in the shown network

• A. $0\text{ A}$
• B. $3\text{ A}$
• C. $1\text{ A}$
• D. $2\text{ A}$

Hint:

Alternatively, look at it via Nodal Analysis. If you assume the top node has a voltage $V_A$, the right branch KVL gives a loop current of zero because the node voltage perfectly matches the balance point of the circuit. Mesh equation addition here is extremely clean and eliminates the target instantly!

Answer 1

The Correct Option is A

Concept: We can determine the exact current values circulating inside the circuit loops using Mesh Current Analysis, which systematically applies Kirchhoff's Voltage Law (KVL) around each closed path loop. Let us define the two independent clockwise loop mesh currents given in the figure:
• Mesh current $i_1$ circulating around the left-hand side loop.
• Mesh current $i_2$ circulating around the right-hand side loop.

Step 1: Writing down the KVL equation for Mesh 1 (Left Loop).
Moving clockwise around the left loop path: $$-36 + 2(i_1) + 12(i_1 - i_2) + 4(i_1) = 0$$ Grouping the coefficient multipliers for $i_1$ and $i_2$: $$(2 + 12 + 4)i_1 - 12i_2 = 36$$ $$18i_1 - 12i_2 = 36$$ We can simplify this by dividing the entire equation by its greatest common divisor, 6: $$3i_1 - 2i_2 = 6 \quad \cdots \text{(Equation 1)}$$

Step 2: Writing down the KVL equation for Mesh 2 (Right Loop).
Moving clockwise around the right loop path: $$12(i_2 - i_1) + 9(i_2) + 24 + 3(i_2) = 0$$ Note that moving clockwise through the $24\text{V}$ DC source goes from the positive terminal to the negative terminal, representing a potential drop of $+24\text{V}$. Grouping the coefficient terms together: $$-12i_1 + (12 + 9 + 3)i_2 = -24$$ $$-12i_1 + 24i_2 = -24$$ We can simplify this by dividing the entire equation by 12: $$-i_1 + 2i_2 = -2 \quad \cdots \text{(Equation 2)}$$

Step 3: Solving the system of simultaneous equations.
We have: 1) 3i_1 - 2i_2 &= 6
2) -i_1 + 2i_2 &= -2 Let us directly add Equation (1) and Equation (2) together to eliminate the $i_2$ term: $$(3i_1 - 2i_2) + (-i_1 + 2i_2) = 6 + (-2)$$ $$2i_1 = 4$$ $$i_1 = \frac{4}{2} = 2\text{ A}$$ Now, substitute this calculated value of $i_1 = 2\text{ A}$ back into Equation (2) to isolate $i_2$: $$-2 + 2i_2 = -2$$ $$2i_2 = -2 + 2$$ $$2i_2 = 0$$ $$i_2 = 0\text{ A}$$ Thus, the current $i_2$ is exactly $0\text{ A}$, which corresponds to Option (A).