Final answer should be 9a: A light bulb is in a circuit with a capacitor. The sw

Question

Final answer should be

9a: A light bulb is in a circuit with a capacitor. The switch is flipped to position A and left there for a long time. Describe what happens to the brightness of the bulb. Now the switch is flipped to position B. Again describe what happens to the brightness of the bulb. (In a class demonstration, we saw that there was a delay for some bulbs to light since they had to heat up sufficiently. Please assume that those delays are negligible here b: Start from the beginning again. The switch is flipped to position A. Determine the voltage across the capacitor 2.5 s later. Use E 50 V, R 250 2 and C 5000 AuF c: At the 2.5 s mark, the switch is suddenly flipped to position B. Determine the voltage across the light bulb 1 s later. d: Verify that the energy stored in the capacitor (at the 2.5 s mark) is all lost in the resistor when the capacitor discharges. e: Redo part c if there was also a 100 S resistor in the top branch of the circuit. BO A

Explanation / Answer

a)

when the switch is flipped to position A, the capacitor is initial uncharged and all the voltage appears across the bulb and bulbs glows to its full. as the capacitor start charging , voltage across it keeps increasing and voltage across the bulb keeps decreasing , hence the brighness of the bulb keeps decreasing and finall the bulb stops glowing when the capacitor gets fully charge and flow of charge stops in the circuit

when the switch is flipped to position B, the capacitor is initial fully charged and all the voltage appears across the capacitor and bulbs has no voltage so it does not glow initially. as the capacitor start discharging , voltage across it keeps decreasing and voltage across the bulb keeps increasing , hence the brighness of the bulb keeps increasing and finall the bulb glows to its full when the capacitor gets fully discharged

b)

T = time constant = RC = 250 x 5000 x 10-6 = 1.25 sec

t = given time = 2.5 sec

Q = charge in the capacitor after t = 2.5 sec = Qo (1 - e-t/T) = (CE) (1 - e-t/T)

Q = (5000 x 10-6) (50) (1 - e-2.5/1.25) = 0.216 C

V = Voltage across the capacitor = Q/C = 0.216/0.005 = 43.2 volts

c)

Q' = = charge on capacitor after 1 sec of discharge

Q' = Q e-t/T

Q' = (0.216) e-1/1.25

Q' = 0.097

V' = Voltage across capacitor = Voltage across the bulb = Q'/C = 0.097/0.005 = 19.4 volts

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