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 determine the root mean square (rms) value of the current when two circuit elements (X and Y) are connected in series, we need to understand the behavior of each element with the given A.C. supply and their phase relationship.

1. When element X is connected to an A.C. supply with peak voltage \(V_{\text{peak}} = 100 \, \text{V}\), it gives a peak current \(I_X = 5 \, \text{A}\), which is in phase with the voltage. This indicates that element X is a resistor.
2. The impedance \(Z_X\) of a resistor is purely real, and it can be calculated as: 
   \(Z_X = \frac{V_{\text{peak}}}{I_X} = \frac{100}{5} = 20 \, \Omega\)
3. For element Y, the peak current \(I_Y = 5 \, \text{A}\) lags behind the voltage by \(\frac{\pi}{2}\), indicating that it is an ideal inductor. In an inductor, current lags behind voltage.
4. Since both elements give the same peak current with the same A.C. supply, the peak current through element Y is also \(5 \, \text{A}\).
5. The impedance of an inductor, \(Z_Y\), can be calculated using the same formula: 
   \(Z_Y = \frac{V_{\text{peak}}}{I_Y} = \frac{100}{5} = 20 \, \Omega\) (This is the inductive reactance \(X_L\)).
6. When X and Y are connected in series, we need to find the total impedance \(Z_{\text{total}}\). Because X (resistor) and Y (inductor) are in series, the total impedance is calculated using Pythagorean theorem: 
   \(Z_{\text{total}} = \sqrt{Z_X^2 + Z_Y^2} = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2} \, \Omega\)
7. Next, we calculate the rms value of the current through this series combination. The peak voltage is 100 V, so the rms voltage \(V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{100}{\sqrt{2}} \, \text{V}\).
8. Using Ohm's law for A.C. circuits: 
   \(I_{\text{rms}} = \frac{V_{\text{rms}}}{Z_{\text{total}}} = \frac{\frac{100}{\sqrt{2}}}{20\sqrt{2}} = \frac{100}{40} = \frac{5}{2} \, \text{A}\)

Therefore, the rms value of the current when X and Y are connected in series and supplied with the same A.C. voltage is \(\frac{5}{2}\) A.