In the circuit below there are two identical bulbs. The switch has been open for

Question

In the circuit below there are two identical bulbs. The switch has been open for a long time and the capacitor is uncharged. The battery is ideal and has a voltage (Vbat) of 12.0 volts.

QUESTION 1: Just after the switch is closed, rank the magnitudes of the four voltages Vbat, VA, VB, and Vcap. Choose the correct option

A: VA = VB > Vcap > Vbat

B: Vbat > VA = VB > Vcap

C:Vbat = VA = VB > Vcap

D: Vbat = VA = VB = Vcap

E: Vbat = VA = Vcap > VB

QUESTION 2: After the switch has been closed for a long time, rank the magnitudes of the currents ibat, iA, and iB. Choose the correct option.

A: iA > iB = ibat

B: ibat = iA = iB

C: ibat > iA = iB

D: ibat = iA > iB

QUESTION 3: Now the Switch is Opened.

Just after the switch is opened, what is the magnitude of the voltage across bulb A?

Explanation / Answer

Just after switch is closed ,Capacitor acts as short circuit ,So Voltage across capacitor is zero i.e

Vc=0 Volts

Since Bulb A and Bulb B are identical

VA=VB=Vbat

So

Vbat=VA=VB>Vc

2.

After switch has been closed for long time ,capacitor acts as open circuit ,so there is no current through capacitor and Bulb B i.e

Icap=0 A

IB=0 A

an Current through IA and battery are equal

IA=Ibat=Vbat/RA

so

Ibat=IA>IB

c)

Voltage across bulb A is

VA=12 Volts

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