When an alternating voltage of `220 V` is applied across a device `P`, a current of `0.25 A` flows through the circuit and it leads the applied voltage by a angle `pi/2` radian. When the same voltage source is connected across another device `Q`, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of `P` and `Q`?

Options

A. `1/(4sqrt2)` A lagging in phase by `pi/4` with voltage
B. `1/(4sqrt2)` A leading in phase by `pi/4` with voltage
C. `1/sqrt2` A leading in phase by `pi/4` with voltage
D. `1/(4sqrt2)` A leading in phase in `pi/2` with voltage

Explanation

To solve the problem, we need to analyze the two devices, P and Q, separately and then find the current when they are connected in series.

### Step 1: Analyze Device P
When an alternating voltage of 220 V is applied across device P, a current of 0.25 A flows through the circuit, leading the applied voltage by an angle of \( \frac{\pi}{2} \) radians. This indicates that device P behaves like a capacitor.

**Hint:** Remember that in a purely capacitive circuit, the current leads the voltage by \( \frac{\pi}{2} \) radians.

### Step 2: Calculate the Impedance of Device P
For device P, we can calculate the capacitive reactance \( X_C \) using the formula:
\[ X_C = \frac{V_{rms}}{I_{rms}} = \frac{220 \, \text{V}}{0.25 \, \text{A}} = 880 \, \Omega \]

**Hint:** Use the formula for reactance to find the impedance in a capacitive circuit.

### Step 3: Analyze Device Q
When the same voltage source is connected across device Q, the same current of 0.25 A is observed, but this time the current is in phase with the applied voltage. This indicates that device Q behaves like a resistor.

**Hint:** In a purely resistive circuit, the current is in phase with the voltage.

### Step 4: Calculate the Resistance of Device Q
For device Q, we can calculate the resistance \( R \) using the same formula:
\[ R = \frac{V_{rms}}{I_{rms}} = \frac{220 \, \text{V}}{0.25 \, \text{A}} = 880 \, \Omega \]

**Hint:** The resistance can be calculated in the same way as the reactance for device P.

### Step 5: Calculate the Total Impedance in Series
When devices P and Q are connected in series, the total impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{(880 \, \Omega)^2 + (880 \, \Omega)^2} = \sqrt{2 \times (880)^2} = 880\sqrt{2} \, \Omega \]

**Hint:** Use the Pythagorean theorem to calculate the total impedance in a series circuit with resistive and reactive components.

### Step 6: Calculate the Net Current in the Series Circuit
Now, we can find the net current \( I_{net} \) when the same voltage source is connected across the series combination of devices P and Q:
\[ I_{net} = \frac{V_{rms}}{Z} = \frac{220 \, \text{V}}{880\sqrt{2} \, \Omega} = \frac{1}{4\sqrt{2}} \, \text{A} \]

**Hint:** Use Ohm's law to find the current in the series circuit.

### Step 7: Determine the Phase Angle
To find the phase angle \( \phi \) of the net current with respect to the applied voltage, we use:
\[ \cos \phi = \frac{R}{Z} = \frac{880}{880\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Thus, \( \phi = \frac{\pi}{4} \) radians.

**Hint:** The phase angle can be found using the cosine of the ratio of resistance to total impedance.

### Final Answer
The current when the same source is connected across a series combination of devices P and Q is:
\[ I_{net} = \frac{1}{4\sqrt{2}} \, \text{A} \]
And the phase angle is \( \frac{\pi}{4} \) radians.