Question

A single phase AC distributor has a total resistance of \(0.2\ \Omega\) and a reactance of \(0.3\ \Omega\). If the load current is \(100\ \text{A}\) at \(0.8\) p.f. lagging, the approximate voltage drop is

• A. \(16\ \text{V}\)
• B. \(34\ \text{V}\)
• C. \(50\ \text{V}\)
• D. \(25\ \text{V}\)

Hint:

For lagging power factor, AC voltage drop is \(I(R\cos\phi+X\sin\phi)\).

Answer 1

The Correct Option is B

Concept: Approximate voltage drop in an AC distributor is: \[ V_d=I(R\cos\phi+X\sin\phi) \] for lagging power factor.

Step 1: Given: \[ I=100\ \text{A} \] \[ R=0.2\ \Omega \] \[ X=0.3\ \Omega \] \[ \cos\phi=0.8 \]

Step 2: Find \(\sin\phi\). \[ \sin\phi=\sqrt{1-\cos^2\phi} \] \[ \sin\phi=\sqrt{1-0.8^2} \] \[ \sin\phi=\sqrt{1-0.64}=0.6 \]

Step 3: Apply voltage drop formula. \[ V_d=100[(0.2)(0.8)+(0.3)(0.6)] \] \[ V_d=100[0.16+0.18] \] \[ V_d=100(0.34) \] \[ V_d=34\ \text{V} \] Therefore: \[ \boxed{34\ \text{V}} \]