A pure resistive circuit element `X` when connected to an sinusoidal `AC` supply peak voltage `200 V` gives a peak current of `5A` 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 currrent but the current lags behind by `90^(0)`. If the series combination of `X` and `Y` is connected to the same supply. the rms value of current is

Options

A. a.`(10)/(sqrt2)amp`
B. b.`(5)/(sqrt2)amp`
C. c.`((5)/(2))amp`
D. d.5amp

Answer

To solve the problem step by step, we will analyze the given information and apply relevant formulas.

### Step 1: Determine the resistance of circuit element X
Given:
- Peak voltage \( V_0 = 200 \, V \)
- Peak current \( I_0 = 5 \, A \)

Since element X is purely resistive, we can use Ohm's law to find the resistance \( R \):

\[ R = \frac{V_0}{I_0} = \frac{200 \, V}{5 \, A} = 40 \, \Omega \]

### Step 2: Analyze circuit element Y
Element Y has a current that lags the voltage by \( 90^\circ \). This indicates that Y is an inductor. The inductive reactance \( X_L \) can be calculated using the same peak current:

\[ X_L = \frac{V_0}{I_0} = \frac{200 \, V}{5 \, A} = 40 \, \Omega \]

### Step 3: Calculate the total impedance \( Z \) of the series combination of X and Y
In a series circuit, the total impedance \( Z \) is given by:

\[ Z = \sqrt{R^2 + X_L^2} \]

Substituting the values we found:

\[ Z = \sqrt{(40 \, \Omega)^2 + (40 \, \Omega)^2} = \sqrt{1600 + 1600} = \sqrt{3200} = 40\sqrt{2} \, \Omega \]

### Step 4: Calculate the RMS voltage \( V_{rms} \)
The RMS voltage \( V_{rms} \) can be calculated from the peak voltage:

\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200 \, V}{\sqrt{2}} = 100\sqrt{2} \, V \]

### Step 5: Calculate the RMS current \( I_{rms} \)
Using Ohm's law for AC circuits, the RMS current can be calculated as:

\[ I_{rms} = \frac{V_{rms}}{Z} \]

Substituting the values:

\[ I_{rms} = \frac{100\sqrt{2} \, V}{40\sqrt{2} \, \Omega} = \frac{100}{40} = 2.5 \, A \]

### Final Answer
The RMS value of the current \( I_{rms} \) is:

\[ \boxed{2.5 \, A} \]