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 10.0 volts.

Just after the switch is closed, rank the magnitudes of the four voltages Vbat, VA, VB, and Vcap.

Vbat = VA = VB = VcapVbat = VA = Vcap > VB     VA = VB > Vcap > VbatVbat = VA = VB > VcapVbat > VA = VB > Vcap

After the switch has been closed for a long time, rank the magnitudes of the currents ibat, iA, and iB.

ibat = iA = iBibat = iA > iB     iA > iB = ibatibat > iA = iB

Then the switch is opened.

Just after the switch is opened, the magnitude of the voltage across bulb A is: V

Explanation / Answer

when the switch is opened for a long time

voltage source is not connected to bulbs as well as capacitor

it means that

bulbs dont have any voltage to develop brightness

capacitor is uncharged

Vcap =0

when the switch is connected

circuit will be formed

Vbat =10

voltage across bulb A is absolutely

=10V

capacitor does not allow sudden changes in voltage

so apply kvl in second mesh

-VA+VC+VB=0

Vbat = VA = VB >Vcap when the Just after the switch is closed

After the switch has been closed for a long time:

ibat > iA = iB

sine capcitor will act as shortcircuit and the voltage diveded equally between bulb A and bulb B

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