A bulb of 100 W is connected in parallel to an ideal inductance of 1H. This arrangement is connected to a 90 V barrery through a switch. On pressing the switch the

Options

A. Bulb does not glow
B. Bulb glows
C. Bulb glows after a short time and then continues to glow
D. Bulb glows for a short time and stops glowing

Explanation

To solve the problem, we need to analyze the circuit consisting of a 100 W bulb and an ideal inductor connected in parallel to a 90 V battery. We will determine what happens when the switch is pressed.

### Step-by-Step Solution:

1. **Identify the Components**:
- We have a 100 W bulb connected in parallel with an ideal inductor of 1 H.
- The circuit is powered by a 90 V battery.

2. **Calculate the Resistance of the Bulb**:
- The power (P) of the bulb is given as 100 W.
- The voltage (V) across the bulb is 90 V.
- Using the formula for power:
\[ P = \frac{V^2}{R} \]
Rearranging gives:
\[ R = \frac{V^2}{P} \]
- Substituting the values:
\[ R = \frac{90^2}{100} = \frac{8100}{100} = 81 \, \Omega \]

3. **Understand the Behavior of the Inductor**:
- The inductor is ideal and has a reactance (X_L) that depends on frequency (f). In a DC circuit, the frequency is 0, thus:
\[ X_L = 2\pi f L = 2\pi \cdot 0 \cdot 1 = 0 \, \Omega \]
- This means that the inductor behaves like a short circuit when the switch is pressed.

4. **Current Path Analysis**:
- Since the inductor has zero resistance, the current will prefer to flow through the inductor rather than the bulb.
- As a result, the current bypasses the bulb.

5. **Conclusion**:
- Since the current does not flow through the bulb, it will not glow when the switch is pressed.

### Final Answer:
The bulb does not glow when the switch is pressed.