Question

A single-phase AC voltage source has 200V (RMS) and a system connected consumes an active power of 300 Watts. What is the reactive power consumed by the system if 2.5A (RMS) current is drawn?

• A. 200 VAR
• B. 300 VAR
• C. 400 VAR
• D. 500 VAR

Hint:

To calculate the reactive power, use the formula \( S = \sqrt{P^2 + Q^2} \), where \(P\) is the active power and \(S\) is the apparent power, which can be found from \( S = V_{\text{rms}} \times I_{\text{rms}} \).

Answer 1

The Correct Option is C

The active power (\(P\)) and reactive power (\(Q\)) are related to the apparent power (\(S\)) by the following equations: \[ S = \sqrt{P^2 + Q^2} \] \[ S = V_{\text{rms}} \times I_{\text{rms}} \] Where: - \(S\) = Apparent power
- \(P\) = Active power
- \(Q\) = Reactive power
- \(V_{\text{rms}}\) = RMS voltage
- \(I_{\text{rms}}\) = RMS current
We are given: - \(V_{\text{rms}} = 200 \, \text{V}\)
- \(I_{\text{rms}} = 2.5 \, \text{A}\)
- \(P = 300 \, \text{W}\)
First, calculate the apparent power \(S\): \[ S = V_{\text{rms}} \times I_{\text{rms}} = 200 \times 2.5 = 500 \, \text{VA} \] Next, use the relationship \(S = \sqrt{P^2 + Q^2}\) to find \(Q\): \[ 500 = \sqrt{300^2 + Q^2} \] Square both sides: \[ 250000 = 90000 + Q^2 \] Solve for \(Q^2\): \[ Q^2 = 250000 - 90000 = 160000 \] Take the square root: \[ Q = \sqrt{160000} = 400 \, \text{VAR} \] Thus, the reactive power is 400 VAR.