Electrical Engineering - Parallel Circuits - Discussion

(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.

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.

The remaining two lights glow same as earlier. Because in parallel circuit voltage is same. Since each bulb is receiving some currents separately, These will glow independently. Suppose third branch line is cutting, third bulb will not glow. The remaining two will glow continuously because these two bulb is receiving separate current from source.

Answer can be (C) also because when the 3rd bulb will blow out, that means NO resistance, then it would be a short-circuit path for the current and so, all the current will flow through the short-circuited branch and other two bulbs would not receive any current at all.

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

Option D, is correct.

Because the resistance of remaining two bulbs remain same and voltage of bulb remains same. So current also remains same. The power consumed by the bulb remains same.

The voltage is same in all the 3 branches in case of parallel circuit. So other two bulbs will be continue to glow with same brightness as before.

The current flowing to the third bulb will be diverted to the other two bulbs, so the other two bulbs will glow brighter.

Anand your answer is wrong. In series ckt current is same, if parallel voltage will be the same.

Absolutely, @Prakash is wrong, if any bulb is blow out, that means is will be opened circuit.