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

Answer: (D)

$\frac{5}{2}$

Answer 2

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
irms=$\frac{1}{√2}i$
=$\frac{5}{2}$
So, the correct option is (D): $\frac{5}{2}$