Question

An AC voltage controller reduces the RMS voltage across a resistive load to 0.5 of its original value. What is the reduction in power?

• A. 50%
• B. 75%
• C. 25%
• D. No change

Hint:

Power scales quadratically with voltage (\(P \propto V^2\)).
If a voltage is multiplied by a factor \(x\), the power is multiplied by \(x^2\).
Here, \(x = 0.5 \implies x^2 = 0.25\). Thus, the remaining power is 25%, meaning the reduction is \(100\% - 25\% = 75\%\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the percentage reduction in power consumed by a resistive load when an AC voltage controller reduces the RMS voltage across it to 0.5 (or half) of its original value.

Step 2: Key Formula or Approach:
For a resistive load of resistance \(R\), the power \(P\) dissipated in terms of the RMS voltage \(V_{\text{RMS}}\) is given by:
\[ P = \frac{V_{\text{RMS}}^2}{R} \] This shows that power is directly proportional to the square of the RMS voltage:
\[ P \propto V_{\text{RMS}}^2 \]

Step 3: Detailed Explanation:

• Let the initial RMS voltage be \(V_1\) and the initial power be \(P_1 = \frac{V_1^2}{R}\).

• The new RMS voltage is \(V_2 = 0.5 V_1\).

• The new power \(P_2\) is:
\[ P_2 = \frac{V_2^2}{R} = \frac{(0.5 V_1)^2}{R} = 0.25 \frac{V_1^2}{R} = 0.25 P_1 \]
• The reduction in power (\(\Delta P\)) is:
\[ \Delta P = P_1 - P_2 = P_1 - 0.25 P_1 = 0.75 P_1 \]
• To express this as a percentage reduction:
\[ \text{Percentage Reduction} = \frac{\Delta P}{P_1} \times 100\% = 0.75 \times 100\% = 75\% \]
• Thus, reducing the RMS voltage by 50% results in a 75% reduction in power.

Step 4: Final Answer:
The reduction in power is 75%.