A pure resistive circuit element X when connected to an ac supply of peak voltage 200 V gives a 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 suply, what will be the rms value of current?

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters for the resistive circuit element X
Given:
- Peak voltage \( V_0 = 200 \, V \)
- Peak current \( I_0 = 5 \, A \)

Since X is a pure resistive circuit, the relationship between voltage and current is given by Ohm's law:
\[ I_0 = \frac{V_0}{R} \]
From this, we can find the resistance \( R \):
\[ R = \frac{V_0}{I_0} = \frac{200 \, V}{5 \, A} = 40 \, \Omega \]

### Step 2: Identify the parameters for the inductive circuit element Y
For circuit element Y, the peak current also remains the same:
- Peak current \( I_0 = 5 \, A \)
- The current lags the voltage by \( 90^\circ \), indicating it is purely inductive.

The inductive reactance \( X_L \) can be calculated in a similar way:
\[ X_L = \frac{V_0}{I_0} = \frac{200 \, V}{5 \, A} = 40 \, \Omega \]

### Step 3: Calculate the total impedance of the series combination of X and Y
The total impedance \( Z \) in a series circuit with resistance \( R \) and inductive reactance \( X_L \) 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 value of the current
The RMS voltage \( V_{rms} \) is given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200 \, V}{\sqrt{2}} = 100\sqrt{2} \, V \]

Now, we can find the RMS current \( I_{rms} \) using Ohm's law:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
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 when the series combination of X and Y is connected to the same AC supply is:
\[ I_{rms} = 2.5 \, A \]