A pure resistive circuit element `X` when connected peak current of `5 A` which is in phase with the voltage. A second circuit element `Y`, when connected to the same `AC` supply also gives the same value of peak current but the current lags behind by `90^(@)`. If the series combination of `X` and `Y` is connected to the same supply, what will be the rms value of current ?

Options

A. `(10)/(sqrt2) A`
B. `(5)/(sqrt2) A`
C. `(5//2) A`
D. `5 A`

Explanation

To solve the problem step by step, we will analyze the given circuit elements and their characteristics.

### Step 1: Identify the Circuit Elements
- Circuit element `X` is a pure resistive element, which means the current is in phase with the voltage.
- Circuit element `Y` is an inductor, where the current lags behind the voltage by 90 degrees.

### Step 2: Determine the Peak Current and Voltage
- The peak current through both elements is given as \( I_0 = 5 \, A \).
- The peak voltage across the circuit can be calculated using Ohm's law for the resistive element \( X \).

### Step 3: Calculate the Resistance of Element X
Using the relationship between peak voltage and peak current in a resistive circuit:
\[ V_0 = I_0 \times R \]
Given that \( I_0 = 5 \, A \), we can express the resistance \( R \) as:
\[ R = \frac{V_0}{I_0} \]
Assuming a peak voltage \( V_0 = 200 \, V \):
\[ R = \frac{200}{5} = 40 \, \Omega \]

### Step 4: Calculate the Inductive Reactance of Element Y
For the inductor \( Y \), the inductive reactance \( X_L \) can be calculated using the same peak voltage and current:
\[ X_L = \frac{V_0}{I_0} \]
Thus,
\[ X_L = \frac{200}{5} = 40 \, \Omega \]

### Step 5: Calculate the Total Impedance in Series
When elements \( X \) and \( Y \) are connected in series, the total impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{40^2 + 40^2} = \sqrt{1600 + 1600} = \sqrt{3200} = 40\sqrt{2} \, \Omega \]

### Step 6: Calculate the RMS Voltage
The RMS voltage \( V_{RMS} \) can be calculated from the peak voltage:
\[ V_{RMS} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, V \]

### Step 7: Calculate the RMS Current
Using Ohm's law for AC circuits, the RMS current \( I_{RMS} \) can be calculated as:
\[ I_{RMS} = \frac{V_{RMS}}{Z} \]
Substituting the values:
\[ I_{RMS} = \frac{100\sqrt{2}}{40\sqrt{2}} = \frac{100}{40} = 2.5 \, A \]

### Final Answer
Thus, the RMS value of the current when the series combination of \( X \) and \( Y \) is connected to the same AC supply is:
\[ \boxed{2.5 \, A} \]