Electrical Engineering - Parallel Circuits - Discussion

You are assuming a constant current source and no load on each branch. In that case yes your answer would be right.

However, consider this:

The current through any one resistor (or branch of the circuit) will be I = V/R = 10/1000 = 0.01I = V/R = 10/1000 = 0.01.

The total current drawn from the power supply will, of course, be 4x that since there are 4 branches, all at 10mA.

Now, remove one branch, and what happens?
Each branch still has I = V/R = 10/1000 = 0.01, since that hasn't changed at all.
What has changed though is the total current?

Since there are now only 3 branches the total current drops to 30mA, as that is 3 x 0.01.
The only ways the current through any branch could change is if a) the resistance in that one branch changes, or b) the voltage of the power source changes.

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.

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.

In parallel branch circuit, there is equal voltage drop in each branch. Since equal current flows in each branch. Hence there is equal resistance present in each branch.

I1*R1 = I2*R2.
60*(R/5) = I2*(R/4).
I2 = 48.

Current in each branch will be 48/4 = 12.

This is tricky q, they are asking what is the current in other 4 branch if the resistance is opened.

So as per above the answer is 12 but logically its 15 in each branch.

(ie) 12 + (12/4) = 12 + 3 = 15.

I agree with @Ashwath.

If voltage remains same means then there will be same current in circuit.

If one branch is removed means total current is distributed among four resistors.

So 60ma/4 = 15ma.

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.

Total current in this circuit must be = 12+12+12+12+12 = 60 mA.

So, if we remove one branch then total current must be same.

Therefore answer must be = 60/4 =15 mA.

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.

For five parallel branch total current is 60mA
If one branch is open then the total current become 60-12=48mA
So current in other branch 48/4=12mA