Question

An a.c. source of 200 V rms supplies active power of 600 W and reactive power of 800 VAR. The rms current drawn from the source is ____.

• A. 10 A
• B. 5 A
• C. 3.75 A
• D. 2.5 A

Hint:

The values 600, 800, and 1000 form a 6:8:10 (or 3:4:5) Pythagorean triple. Recognizing these triples allows you to find the hypotenuse (Apparent Power) instantly without doing long calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In AC circuits, power is composed of three components: Active Power ($P$), Reactive Power ($Q$), and Apparent Power ($S$). These three form a right-angled triangle known as the Power Triangle.

Step 2: Key Formula or Approach:
The relationship between the powers is: \[ S = \sqrt{P^2 + Q^2} \] And the Apparent Power is also calculated as: \[ S = V_{\text{rms}} \times I_{\text{rms}} \]

Step 3: Detailed Explanation:
1. Calculate Apparent Power ($S$): Given $P = 600\text{ W}$ and $Q = 800\text{ VAR}$. \[ S = \sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} \] \[ S = \sqrt{1000000} = 1000\text{ VA} \] 2. Calculate RMS Current ($I_{\text{rms}}$): Given $V_{\text{rms}} = 200\text{ V}$. Using $S = V_{\text{rms}} \times I_{\text{rms}}$: \[ 1000 = 200 \times I_{\text{rms}} \] \[ I_{\text{rms}} = \frac{1000}{200} = 5\text{ A} \]

Step 4: Final Answer:
The rms current drawn from the source is 5 A.

Answer 2

Instead of computing the apparent power directly, we can test each candidate current against the given active power \( P = 600 \) W and reactive power \( Q = 800 \) VAR, and check whether it produces a mathematically consistent power triangle, where \( \cos^2\phi + \sin^2\phi = 1 \).

1. 10 A: Apparent power would be \( S = VI = 200 \times 10 = 2000 \) VA. This implies \( \cos\phi = \frac{P}{S} = \frac{600}{2000} = 0.3 \) and \( \sin\phi = \frac{Q}{S} = \frac{800}{2000} = 0.4 \). Checking, \( 0.3^2 + 0.4^2 = 0.09 + 0.16 = 0.25 \), which is not \( 1 \), so this current is inconsistent with the given powers.
2. 5 A: Apparent power would be \( S = 200 \times 5 = 1000 \) VA, giving \( \cos\phi = \frac{600}{1000} = 0.6 \) and \( \sin\phi = \frac{800}{1000} = 0.8 \). Checking, \( 0.6^2 + 0.8^2 = 0.36 + 0.64 = 1 \), exactly satisfying the identity, so this current is fully consistent.
3. 3.75 A: Apparent power would be \( S = 200 \times 3.75 = 750 \) VA, giving \( \cos\phi = \frac{600}{750} = 0.8 \) but \( \sin\phi = \frac{800}{750} \approx 1.067 \), which exceeds 1. Since \( \sin\phi \) can never exceed 1, this current is physically impossible for the given powers.
4. 2.5 A: Apparent power would be \( S = 200 \times 2.5 = 500 \) VA, giving \( \cos\phi = \frac{600}{500} = 1.2 \), which also exceeds 1 and is impossible for the same reason.

Only 5 A satisfies the fundamental identity \( \cos^2\phi + \sin^2\phi = 1 \) exactly, confirming it as the physically valid current.

Therefore, the correct answer is 5 A.