# Electrical Engineering - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 25)

25.

In a five-branch parallel circuit, there are 12 mA of current in each branch. If one of the branches opens, the current in each of the other four branches is

Discussion:

22 comments Page 1 of 3.
Aizaz haider said:
4 years ago

@Piyush Kanani.

Thank you for explaining the right answer.

Thank you for explaining the right answer.

Bhaskar said:
4 years ago

Yes, I agree with you@Ashwath.

Piyush Kanani said:
5 years ago

Imagine system voltage = Y.

The resistance of each branch = X.

During the 5 branches system total resistance = X/5.

So, V=IR= 60*(X/5) = 12X.

So, Y=12X.

NOW, during 4 branches system total resistance = X/4.

So, V = IR= I*(X/4).

I = V/(X/4).

= 12X/(X/4).

= 48.

Which divided into 4 parallel branches so par branch 12.

The resistance of each branch = X.

During the 5 branches system total resistance = X/5.

So, V=IR= 60*(X/5) = 12X.

So, Y=12X.

NOW, during 4 branches system total resistance = X/4.

So, V = IR= I*(X/4).

I = V/(X/4).

= 12X/(X/4).

= 48.

Which divided into 4 parallel branches so par branch 12.

Abhijith said:
5 years ago

There are 5 branches, current in each branch is 12, so total current = i1 + i2 + i3 + i4 + i5 = 60.

Then one branch is an open circuit, that means there is no current in that branch, take i2 = 0,

(if i2 brach taken as opn) so total current will decrease, i1 + 0 + i3 + i4 + i5 = 48 (12 * 4).

So, current in each branch is 12 (48Ã·4), there are equal resistance in each branch.

Then one branch is an open circuit, that means there is no current in that branch, take i2 = 0,

(if i2 brach taken as opn) so total current will decrease, i1 + 0 + i3 + i4 + i5 = 48 (12 * 4).

So, current in each branch is 12 (48Ã·4), there are equal resistance in each branch.

(1)

Tanim Hossain said:
6 years ago

@All.'

The Answer B is correct, because the question is (current in each of the other four branches).

The Answer B is correct, because the question is (current in each of the other four branches).

Senthamizh said:
6 years ago

The question is about other branch current. Not total current. So it will remain unaffected.

(1)

Senthamizh said:
6 years ago

Yes, correct @Amit.

Bibek said:
6 years ago

It should be 15. Because no current flows in open circuit.

Waryam said:
7 years ago

Current through 5 branches = 60mA.

If one branch is open then resistance = 12+12+12+12+R= 60.

i.e 48+R= 60.

& R=60-48=12.

If one branch is open then resistance = 12+12+12+12+R= 60.

i.e 48+R= 60.

& R=60-48=12.

(1)

Pawan Bhakuni said:
7 years ago

12A is wrong. If we connect voltage source then 12A is correct but if current source is connected it would be "increases".

Nothing is mentioned about source.

Nothing is mentioned about source.

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