The circuit at right contains a battery, a bulb, a switch, and a capacitor. The

Question

The circuit at right contains a battery, a bulb, a switch, and a capacitor. The capacitor is initially uncharged. a. Describe the behavior of the bulb in the two situations below. i. The switch is first moved to position I. Describe the behavior of the bulb from just after the switch is closed until a long time later. Explain. ii. The switch is now moved to position 2. Describe the behavior of the bulb from just after the switch is closed until a long time later. Explain your reasoning. b. A second identical bulb is now added to the circuit as shown. The capacitor is discharged. i. The switch is now moved to position I. Describe the behavior of bulbs B and C from just after the switch is closed until a long time later. Explain. How does the initial brightness of bulb C compare to the initial brightness of bulb A in question i of part a? Explain your reasoning. A long time after the switch is closed, is the absolute value of the potential difference across the capacitor greater than, less than, or equal to the absolute value of the potential difference across the battery? Explain.

Explanation / Answer

1 (i) We will consider bulb A as a resistance R, so it will be like RC circuit.

when the switch is moved to position 1, current will flow the circuit and initial current will be V/R, where V is potential of battery.

but as soon as current will flow through the circuit, capacitor will start charging with a certain time constant and voltage across capacitor will increas. as the capacitor is being charged up, current will be decreasing.

so initially current flowing through the bulb will be high and as the capacitor is being charged up current will be decreasing, so initially bulb will glow brighter, and slowly slowy intensity will decrease and after a long time (infinite) bulb will stop glowing.

(ii) when the swich is moved to position 2. there will be a potential difference due to charged capacitor. capacitor will start discharging and current will flow through the circuit. because of this current bulb will glow initially, but the voltage across the capacitor Vc will decrease exponentially with respect to time. So the current flowing through bulb will also decrease exponentially as given by I =Vc/R. So the intensity of the bulb will keep decreasing with time and after a long time (infinite) bulb will stop glowing.

b(i) when switch is moved to position 1, bulb B and bulb C will behave same as shown in a(i). both the bulbs will glow brighter initially and intensity will decrease slowly. after a long time (infinite) both the bulb will stop glowing.

The only difference will be the intensity of the bulbs. as this time we have two identical bulbs connected in series. if we consider resistance of a bulb to be R then the total resistance of the circuit will be R+R =2R, so the current flowing through the circuit at the initial stage will be I=V/2R. so in this case the current flowing through the circuit will be half compared to the situation a(i). So the initial brightness of the bulb C will be half of the initial brightness of bulb A.

After a long time till the potential difference across capacitor and absolute potential difference of battery is different, there will be flow of current from higher voltage to lower voltage, so in equilibrium condition the potential difference across capacitor and absolute potential difference of battery will be same.

Thanks and please let me know if you have any further query

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