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 2 of 3.
Balgopal said:
8 years ago
Simply think like a household wiring, suppose we are making a load switched off (open ckt). Then net load (A) will be decreased. New net load /4 = 12.
Sushil ram said:
8 years ago
Well said @Prashant Jayant.
Pawan Bhakuni said:
8 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.
Waryam said:
8 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)
Bibek said:
7 years ago
It should be 15. Because no current flows in open circuit.
Senthamizh said:
7 years ago
Yes, correct @Amit.
Senthamizh said:
7 years ago
The question is about other branch current. Not total current. So it will remain unaffected.
(1)
Tanim Hossain said:
7 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).
Abhijith said:
7 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 (48÷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 (48÷4), there are equal resistance in each branch.
(1)
Piyush Kanani said:
6 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.
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