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

To solve this problem, we first need to understand the behavior of the two elements X and Y when connected to an AC supply:

Element X gives a peak current of 5 A which is in phase with the voltage, meaning X behaves purely resistively. Therefore, the impedance of X is given by: 

\(R_X = \frac{V_{\text{peak}}}{I_{\text{peak}}} = \frac{100}{5} = 20 \, \Omega\).

Element Y gives a peak current of 5 A which lags the voltage by \(\frac{π}{2}\), indicating that Y behaves as a pure inductor. The impedance of Y (inductive reactance \((X_L)\)) is:

\(X_L = \frac{V_{\text{peak}}}{I_{\text{peak}}} = \frac{100}{5} = 20 \, \Omega\).

Now, when X and Y are connected in series, the total impedance \((Z_{\text{total}})\) is the vector sum of the resistive and inductive reactances:

\(Z_{\text{total}} = \sqrt{R_X^2 + X_L^2} = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \Omega\).

The peak current \((I_{\text{peak, total}})\) for the series circuit is given by:

\(I_{\text{peak, total}} = \frac{V_{\text{peak}}}{Z_{\text{total}}} = \frac{100}{20\sqrt{2}} = \frac{5}{\sqrt{2}} \, \text{A}\).

The RMS (root mean square) value of the current is:

\(I_{\text{RMS}} = \frac{I_{\text{peak, total}}}{\sqrt{2}} = \frac{5/\sqrt{2}}{\sqrt{2}} = \frac{5}{2} \, \text{A}\).

Thus, the correct answer is \(\frac{5}{2}\) amperes.

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