Two bulbs and a capacitor are connected to a battery as shown below. The capacit

Question

Two bulbs and a capacitor are connected to a battery as shown below. The capacitor is initially uncharged.

a) Just after the switch is closed:

(i) What is the potential difference across bulb A, across bulb B, across the capacitor, and across the battery? Explain.

(ii) Rank the currents through bulb A, bulb B, the capacitor, and the battery. Explain.

b) A long time after the switch is closed:

(i) Rank the currents through bulb A, bulb B, the capacitor, and the battery. Explain.

(ii) What is the potential difference across bulb A, across bulb B, across the capacitor, and across the battery? Explain.

c) Summarize your results by describing the behavior of bulb A and of bulb B from just after the switch is closed until a long time later.

Explanation / Answer

a)

Just after the switch is closed,

(i)

the capacitor will be short cicuited.

So, the bulb B also gets shorted and in the circuit only bulb A remains.

So, voltage across bulb A, VA = emf of the battery (as there is only bulb A in the circuit so all the voltage is across bulb A)

voltage across bulb B, VB = 0 <--- as bulb B gets shorted out

voltage across capacitor, VC = 0 < ----- as it gets shorted out

(ii)

current through bulb A , IA = E/R

where E = emf of battery

R = resistance of bulb

current of bulb B, IB = 0

current through capacitor = 0

current through battery = E/R

b)

After a long time, the capacitor acts as an open circuit and the whole current flows through the two bulbs only.

(ii)

So, Voltage across each bulb = E/2 <----- assuming both bulbs are identical. So voltage will be shared equally

Voltage across capacitor = E/2 <----- voltage across capacitor = voltagae across bulb B

voltage across battery = E <---- assuming the battery is ideal

(i)

current across each bulb = E/2R

where 2R = equivalent resistance of circuit

current across capacitor = 0

current in battery = E/2R

c)

Just after the switch is closed, bulb A will glow with greater brightness than when it remains after a long time later.

For bulb B, it will be initially off when the switch was closed thereafter the brightness will increase

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