Topic: Network Theory
The voltage across the resistor R is:
- A. 1 / √3 ∠0° V
- B. 1 / √2 ∠0° V
- C. 1 / 2 ∠45° V
- D. 1 / √2 ∠45° V
Answer: D. 1 / √2 ∠45° V
Explanation: Z_1 = (1 + j1) Ω = √2 ∠45° Ω and Z_2 = (1 - j1) Ω = √2 ∠-45° Ω. Using current division, I_R = [Z_1 / (Z_1 + Z_2)] * I = [√2 ∠45° / (√2 ∠45° + √2 ∠-45°)] * 1 = (1 / √2) ∠45° A. The voltage across R = 1 Ω is V_R = I_R * R = (1 / √2) ∠45° V.