An ideal inductor is in turn put across 220 V, 50 Hz and 220 V, 100 Hz supplies. The current flowing through it in the two cases will be

Options

A. equal
B. different
C. zero
D. infinite

Explanation

To solve the problem of finding the current flowing through an ideal inductor when connected to two different AC supplies (220 V, 50 Hz and 220 V, 100 Hz), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Inductor's Behavior**:
An ideal inductor has a property where the voltage across it (V) and the current through it (I) are related by the inductive reactance (X_L). The relationship can be expressed as:
\[ V = I \cdot X_L \]
where \(X_L\) is the inductive reactance.

2. **Calculate Inductive Reactance**:
The inductive reactance \(X_L\) is given by the formula:
\[ X_L = 2 \pi f L \]
where \(f\) is the frequency in hertz (Hz) and \(L\) is the inductance in henries (H).

3. **Determine Current for Each Frequency**:
Rearranging the formula for current, we have:
\[ I = \frac{V}{X_L} \]
Substituting the expression for \(X_L\):
\[ I = \frac{V}{2 \pi f L} \]

4. **Case 1: For 50 Hz Supply**:
- Given: \(V = 220 \, V\), \(f = 50 \, Hz\)
- The current \(I_1\) can be calculated as:
\[ I_1 = \frac{220}{2 \pi (50) L} = \frac{220}{100 \pi L} \]

5. **Case 2: For 100 Hz Supply**:
- Given: \(V = 220 \, V\), \(f = 100 \, Hz\)
- The current \(I_2\) can be calculated as:
\[ I_2 = \frac{220}{2 \pi (100) L} = \frac{220}{200 \pi L} \]

6. **Compare the Two Currents**:
- From the equations for \(I_1\) and \(I_2\):
\[ I_1 = \frac{220}{100 \pi L} \quad \text{and} \quad I_2 = \frac{220}{200 \pi L} \]
- We can see that:
\[ I_1 = 2 \cdot I_2 \]
This shows that the current at 50 Hz is twice the current at 100 Hz.

### Conclusion:
- The current flowing through the inductor when connected to the 50 Hz supply is greater than the current flowing through the inductor when connected to the 100 Hz supply. Thus, the current decreases as the frequency increases.