Question

Three identical resistors are connected to a lightbulb as shown in the figure. Assume the battery and all wires are ideal. Initially the switch is closed and the bulb is on. What happens to the brightness of the bulb if the switch is opened? ーーー The bulb briefly gets dimmer but then returns to its initial brightness. The bulb gets brighter. The bulb's brightness is unchanged. The bulb gets dimmer. The bulb briefly gets brighter but then returns to its original brightness.

Answer 1

RBulb C M ДАМ A31 B1 let all Resistance be in When switch is close Met Resistance of circuit sue to AB + + + = RAB RAB = 1/2 Net Resistance of circiet +1 # Bulb 38 Roulb. 2 of When Circuit is open, Net Resistance circult tlf R Bulb at Roulb. So, when circutt is open, Resistance decreases. By Ohn's laro V = PIR so, curent through the circuit decream By P= I² Roues Power across bulb decreases.

As key opens, the resistance is parallel is removed and so resistance of the whole circuit increases.

We know by Ohm's law - V=IR . V is the voltage and I is the current flowing through the circuit and R is the resistance of equivalent circuit. As Resistance increases, voltage through battery is fixed , the current through the circuit will decrease and hence less current will flow through the bulb. We know P =(I^2)*R, Power through the bulb decreases . hence it will glow dim. The bulb gets dimmer.