Question

Two 120v light bulbs , one of 25w and another of 200w are connected in series . one bulb burnt out almost instantaneously ?.which one was burnt and why?.

Answer 1

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Initially, lets find the resistance of each one of these bulbs,
For bulb 1 : V = 200V and Power, P = 40 W
P = V2 / R = (200)2 /R = 40
R = 1000 ohms
For bulb 2: V = 200 V and power = 100 W
P = V2 / R = (200)2 /R = 100
R = 400 ohms
In this circuit, the combined resistance = 1400 ohms.
V given  = 420 V
I * R combined= 420
I * 1400 = 420
I = 420/ 1400
= 0.3 A
For bulb 1 :
I2 R = 0.32 (1000)
= 90W when its rated power is 40 W
For bulb 2:
I2 R = 0.32 (400)
= 36 W when its rated power is

In the first case, the bulb illuminates brightly for a while. Eventually because of the increased power, it will tend to heat up and heating causes the resistance to increase. Because of which, the bulb will fuse soon.

The second will work perfectly.