Question

Consider the four circuits shown in the figure, each consisting of a battery, a switch, a light bulb, a resistor and either a capacitor or an inductor. Assume the capacitor has a large capacitance and the inductor has a large inductance but no resistance. The light bulb has high efficiency and glows whenever electric current passes through it.

(i) Describe what the light bulb does in each of circuits (a) through (d) after the switch is closed.

(ii) Describe what the light bulb does in each of circuits (a) through (d) when the switch has been closed for a long time interval and when the switch is opened.

Accepted Answer

When the switch is closed: from the given figure (a):

The light bulb glow brightly right away because the capacitor behaves as a short circuit and maximum current starts flowing in the circuit through bulb. Therefore, the bulb glows with maximum brightness and after some time, the brightness of the bulb starts decreasing because the capacitor gets charged-up and it behaves as open circuit.

When the switch is closed: from the given figure (b)

In the given figure, the inductor is connected in series with the bulb and the inductor behave like an open wire at $t = 0$ and as the plane wire is at $t = \infty$ , so the light bulb gradually gets brighter after some time, rapidly at first and then, more and more slowly.

When the switch is closed: from the given figure (c),

In figure (c), the bulb and capacitor are connected in parallel with the battery. So, the potential difference across the bulb or capacitor is same at $t = 0$ . So, the bulb immediately becomes bright and at $t = \infty$ , capacitor behaves as an open circuit. But there is still same potential difference across the bulb.

When the switch is closed: from the given figure (d),

In figure (d), the bulb and the inductor are connected in parallel with the battery. At $t = 0$ , the inductor behaves as an open circuit. So, the potential difference across the battery is same. Hence, the bulb is immediately bright. But at $t = \infty$ , the inductor behaves as a plane wire and after a long time, short-circuit occurs.

When the switch is thrown open after a long time: for (a) and (b),

Circuit is incomplete after it is opened. So, the bulb will not glow.
For (c) and (d), the circuit is complete after it is open. So, the bulb will start glowing due to the discharging of capacitor or discharging of inductor.

B M Sharma Solutions for Chapter: Inductance, Exercise 2: DPP

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Inductance, Exercise 2: DPP

Questions from B M Sharma Solutions for Chapter: Inductance, Exercise 2: DPP with Hints & Solutions

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