Question

A circuit element X when connected to an a.c. supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by \(\frac{π}{2}\). If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

• A. \(\frac{10}{√2}\)
• B. \(\frac{10}{√2}\)
• C. 5√2
• D. \(\frac{5}{2}\)

Answer 1

The Correct Option is D

R=\(\frac{100}{5}\)=20 Ω
XL=\(\frac{100}{5}\)=20 Ω
When in series
\(z=\sqrt{202+20^2}=20√2 Ω\)
i=\(\frac{100}{z}\)=\(\frac{100}{20√2}\)=\(\frac{5}{√2}\)
Then, the rms value of the current
i_rms=\(\frac{1}{√2}i\)
=\(\frac{5}{2}\)
So, the correct option is (D): \(\frac{5}{2}\)