Electrical Engineering - Parallel Circuits - Discussion

Discussion :: Parallel Circuits - General Questions (Q.No.7)

7. 

Three lights are connected in parallel across a 120 volt source. If one light burns out,

[A]. the remaining two will glow dimmer
[B]. the remaining two will glow brighter
[C]. the remaining two will not light
[D]. the remaining two will glow with the same brightness as before

Answer: Option D

Explanation:

No answer description available for this question.

Anand said: (Jul 26, 2011)  
In series circuit voltage remains same on each circuit, if parallel voltage is different.

Muthumayil said: (Aug 25, 2011)  
Anand your answer is wrong. In series ckt current is same, if parallel voltage will be the same.

Ramu said: (Jun 21, 2012)  
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.

Sushil said: (Oct 31, 2014)  
Theoretically it is correct but practically it is wrong. Practically answer should be (B).

Yathi said: (Feb 21, 2015)  
The current flowing to the third bulb will be diverted to the other two bulbs, so the other two bulbs will glow brighter.

Vageesh said: (May 11, 2015)  
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.

Prakash said: (May 28, 2015)  
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.

Naveen said: (Jun 3, 2015)  
@Prakash you are wrong, if 3rd bulb blows out it will become open circuit.

Siva said: (Dec 15, 2015)  
Absolutely, @Prakash is wrong, if any bulb is blow out, that means is will be opened circuit.

Shashidhra said: (Aug 11, 2016)  
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.

Sharan said: (Sep 9, 2016)  
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.

Prashanth said: (Nov 6, 2016)  
I agree with @Sharan.

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