Electrical Engineering - Parallel Circuits - Discussion

When one bulb fails, then the current flowing in that branch will be zero.

So, now we have only two lamps left in the circuit, thus circuit fetches current which is sufficient to glow those two lamps. So option D is correct.

LET ME GIVE ONE EXAMPLE.

At home, if we turn on more devices then current will be increased and power consumed will also increase. If you turn off any one load then power consumption is reduced,

P = VI.
V remains constant but the current has been reduced.

Here also no need to have same power if current reduces power will also decrease.

I agree with @Sharan.

(D): the remaining two will glow with the same brightness as before.

Reasoning: Before, I divide into 3 bulbs so each gets I/3 But After I divide into 2 So each gets I/2.

Because I/2>I/3, Therefore, Bulbs should glow brighter.
But I total before not equals I total after.
Because I after will decrease due to an increase in resistance.
Hence as a net result, I will remain the same.

Mathematical solution:
Let's assume one 3 Bulb of resistance X ohm is in Parallel.

Before, Burn out:
The Total current flow,
I = V/(R total)
I = 120/(X/3)
I = 3*120/X
I = 360/X.

The current flow though Bulb 1, I1 = (R other / (R bulb1+ R other) ) * I total {using Current division Rule}
I1 = { {1/( (1/X)+(1/X) )} / {X+ (1/ ( (1/X)+(1/X) ) )} } * 360/X.
I1 = {(X/2)/(3X/2)} * 360/X.
I1 = {1/3}* 360/X.
I1 = 120/X.

After, Burn out: The Total current flow, I = V/(R total)
I = 120/(X/2).
I = 2*120/X.
I = 240/X.

The current flow though Bulb 1, I1 = (R other / (R bulb1+ R other) ) * I total {using Current division Rule}
I1 = { {X} / {X+X} } * 240/X.
I1 = {(X)/(2X)} * 240/X.
I1 = {1/2}* 240/X.
I1 = 120/X.
We observed, I through Bulb 1 before = I through Bulb 1 after = 120/X.
Also Voltage before = Voltage after = 120 V.

As Glow depending on V and I.
Because Voltage before = Voltage after && current before = current after.
Therefore, Power before = Power after
Therefore, glow before = glow after

Hence, the remaining two will glow with the same brightness as before.

D is the correct answer. Some are saying the current of third branch will be diverted to other two, but in actual, voltage is same in parrallel, so same current will pass through other two as before

I think the answer should be (B).