Electrical Engineering - Series-Parallel Circuits - Discussion

@Meena.

Exactly your right for same set of resistors. And even we can calculate the same in for different resistors values.

@Meena.

You are wrong in parallel current through all branches is not same :).

You are getting the right answer because all resistors are same value so the currents drawn by them is also same :).

Then you need to do as explained by @Priya.

What about if the resistors connected in parallel are not of same value?

I am 100% agree with the solution given by Meena, it is the simplest & quick way of calculation. The total current flowing out from the first parallel combination will flow in the second combination but will get divided in to five equal parts so each part of current will be i/5th of the combination(2nd) total input current which will be 1/5th i.e..2 or.2*100 = 20%.

Current I flowing in the circuit will be divided into 5 equal parts (as the 5 equal valued resistors are connected in parallel).

So the current in each branch (of resistor) is I/5.

So the percentage is : (I/5) /I*100 = 20%.

From Priya calculations, we get:

Current in circuit = .089 A.

Total resistance in parallel combination of 5 1 Kohm resistors = 200 Ohm.

So, voltage in parallel of 5 1kohm resistor = I x R(total res of paralleled 5 resistors).

= .089 x 200.

= 17.8 V.

For parallel combinations, V = V1 = V2 = V3 = V4.

So. V1 = I1 R1.

I1 = V1/R1 = 17.8/(1x10^3) = 0.0178 A.

For percentage, I1/whole circuit current x 100% = 0.0178/.089 x 100.

= 0.2 x 100.

= 20%.

@Rohith, Am Not getting your Formula, Could you please expand Once again ?

Thanks rohit.

The simplest way is to decide the correct ans. is the total current is passed through five parallel 1k resister equal amount bcoz the same value of resister so 1k current flow=total current (%)/number of resister; 20%.