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?
- \(\frac{10}{√2}\)
- \(\frac{10}{√2}\)
- 5√2
- \(\frac{5}{2}\)
The Correct Option is D
Approach Solution - 1
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.
Approach Solution -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): \(\frac{5}{2}\)
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Concepts Used:
LCR Circuit
An LCR circuit, also known as a resonant circuit, or an RLC circuit, is an electrical circuit consist of an inductor (L), capacitor (C) and resistor (R) connected in series or parallel.
Series LCR circuit
When a constant voltage source is connected across a resistor a current is induced in it. This current has a unique direction and flows from the negative to positive terminal. Magnitude of current remains constant.
Alternating current is the current if the direction of current through this resistor changes periodically. An AC generator or AC dynamo can be used as AC voltage source.
