Calculate the value of inductance of an inductor connected across an alternating power supply of 220 V of frequency 50 Hz such that maximum current flowing through it is 1 A.

Answer

To calculate the value of inductance (L) of an inductor connected across an alternating power supply, we can follow these steps:

### Step 1: Understand the Given Values
- RMS Voltage (Vrms) = 220 V
- Frequency (f) = 50 Hz
- Maximum Current (I0) = 1 A

### Step 2: Calculate the RMS Current (IRMS)
The relationship between the maximum current (I0) and the RMS current (IRMS) for a sinusoidal supply is given by:
\[ IRMS = \frac{I0}{\sqrt{2}} \]
Substituting the value of I0:
\[ IRMS = \frac{1}{\sqrt{2}} \approx 0.707 \, \text{A} \]

### Step 3: Relate Impedance (Z) to Inductive Reactance (XL)
In a pure inductive circuit, the impedance (Z) is equal to the inductive reactance (XL):
\[ Z = XL \]
And the inductive reactance is given by:
\[ XL = L \cdot \omega \]
where \(\omega = 2\pi f\).

### Step 4: Use the Relationship Between Voltage, Current, and Impedance
From Ohm's law for AC circuits, we have:
\[ IRMS = \frac{VRMS}{Z} \]
Substituting for Z:
\[ IRMS = \frac{VRMS}{XL} \quad \Rightarrow \quad XL = \frac{VRMS}{IRMS} \]
Substituting the known values:
\[ XL = \frac{220}{0.707} \approx 311.77 \, \Omega \]

### Step 5: Substitute XL in Terms of L and ω
Now we can express XL in terms of L and ω:
\[ XL = L \cdot \omega = L \cdot (2\pi f) \]
Substituting for ω:
\[ XL = L \cdot (2\pi \cdot 50) \]
Setting the two expressions for XL equal to each other:
\[ L \cdot (2\pi \cdot 50) = 311.77 \]

### Step 6: Solve for L
Now, we can solve for L:
\[ L = \frac{311.77}{2\pi \cdot 50} \]
Calculating the denominator:
\[ 2\pi \cdot 50 \approx 314.16 \]
Now substituting this back:
\[ L \approx \frac{311.77}{314.16} \approx 0.99 \, \text{H} \]

### Step 7: Final Answer
Thus, the inductance of the inductor is approximately:
\[ L \approx 1 \, \text{H} \]