KCET · Physics · Electromagnetic Waves
A multi meter reads a voltage of a certain A.C. source as \( 100 \mathrm{~V} \). What is the peak value of
voltage of A.C. source ?
- A \( 200 \mathrm{~V} \)
- B \( 100 \mathrm{~V} \)
- C \( 141.4 \mathrm{~V} \)
- D \( 400 \mathrm{~V} \)
Answer & Solution
Correct Answer
(C) \( 141.4 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
The root mean square value of voltage, \( v_{\mathrm{rms}} \) and the peak voltage, \( v_{0} \) of an ac source are related as
\( v_{\mathrm{rms}}=\frac{v_{0}}{\sqrt{2}} \)
Given, \( v_{\text {rms }}=100 \mathrm{~V} \)
Therefore,
\( v_{0}=\sqrt{2} v_{r m s}=1.44 \times 100=141.4 \mathrm{~V} \)
Thus, peak value of voltage of ac source is \( 141.4 \mathrm{~V} \).
\( v_{\mathrm{rms}}=\frac{v_{0}}{\sqrt{2}} \)
Given, \( v_{\text {rms }}=100 \mathrm{~V} \)
Therefore,
\( v_{0}=\sqrt{2} v_{r m s}=1.44 \times 100=141.4 \mathrm{~V} \)
Thus, peak value of voltage of ac source is \( 141.4 \mathrm{~V} \).
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