Applying the Kirchhoff's rule to the loop (a) ABCA and (b) BCDEB, two correct equations are given
(A) $8-4I_2-2(I_2 +I_3) - I_1 = 0$
(B) $4 - 2(I_2 +I_3)-4(I_2 +I_3-I_1) = 0$
(C) $4+2(I_2-I_3) - 4(I_2 + I_3 − I_1) = 0$
(D) $8+ 4I_2 + 2(I_2-I_3) - I_1 = 0$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Given circuit:
Apply Kirchhoff’s Voltage Law (KVL) to both loops.
Loop (a): ABCA
Starting from A → B → C → A
• A to B: Current through 4Ω → potential drop = 4I₂
• B to C: Current through 2Ω → potential drop = 2(I₂ + I₃)
• C to A: Current through 1Ω → potential drop = I₁
• Source at A: +8 V (rise)
Applying KVL:
$8 - 4I_2 - 2(I_2 + I_3) - I_1 = 0$
→ This matches option (A).
Loop (b): BCDEB
Starting from B → C → D → E → B
• B to C: 2Ω → drop = 2(I₂ + I₃)
• C to D: 4Ω → drop = 4(I₂ + I₃ - I₁)
• D to E: battery of 4 V (rise = +4 V)
Applying KVL:
$4 - 2(I_2 + I_3) - 4(I_2 + I_3 - I_1) = 0$
→ This matches option (B).
Final Answer:
Correct equations are (A) and (B).