Consider the circuit shown to the right. The light bulbs are identical and initi
ID: 1391678 • Letter: C
Question
Consider the circuit shown to the right. The light bulbs are identical and initially the capacitor is uncharged. (a) The switch is closed at time t 30. Describe the behavior of bulb 1 and bulb 2 from just after the switch is closed to a long time later Include your description a comparison of the brightness of the Explain your reasoning L H bulbs. Bulb 1 Bulb 2. (b) Find the potential difference across the capacitor in terms of the voltage across the battery (VBattery) a very long time after the switch is closed. Explain your reasoning. (c) After the switch has been closed for a very long time, the switch is then reopened. Describe the behavior of bulb 1 and buib 2 from just after the switch is opened to a long time later. Explain your reasoning Bulb 1 Bulb 2Explanation / Answer
(a) Bulb 1;The moment switch is closed, current will start flowing in the circuit from the positive terminal of the battery. The capacitor will be charged and the bulb 1 will be illuminated. This bulb will be bright because It is connected in parallel with the capacitor and will get the same voltage as the capacitor will get and hence more current. With the passage of time, the voltage across the capacitor will increase and hence the brightness of bulb 1 also.
Bulb 2 : will be dimmer than the bulb one because the drop across 2 will be lesser by (Vb - Vc), Vc is the drop across capacitor.This bulb will continue to get dimmer becayse, Vc will keep increasing with passage of time, till it become equal to Vb.
(b)A very long time, after the switch is closed,
Vc = Vb
(c)Bulb 1: Long time after, the capacitor is fully charged and its voltage is Vc = Vb, the bulb in parallel will get the same drop across it and will be illuminating. But after the switch is reopened, teh capacitor begins to discharge and the brightness of the bulb will start diminising and a time will come when Vc tends to 0, bulb 1 will stop illuminating meaning will be off.
Bulb 2, The moment, switch is reopened, bulb 2 will not get any potential and hence no current, so it will stop illuminating immediately after the switch has been reopened.