Electrical Engineering - Series-Parallel Circuits - Discussion

@Uday.

Thanks for the explanation.

Three 10k in series but one 10k in parallel with 20k = 10*20/30 = 6.66 k + 20 k = 26.66 k.

I = 24/26.66 k = 900 uA.

In the given question, R1, R2 and R3 are in series and R4 is in parallel with any one. So for example R4||R3.

Now R1 and R2 in series 10+10 = 20 k and from R3 and R4 = 6.66 k, which is now in series with R1 and R2.

Then Rt is 26.66 k.

To find I = v/r.

I = 24v/26.66k = 0.900 mA = 900 uA.

R = R1+R2+R3.
R = 10+10+6.66666 = 26.66666 K ohm.

1/R3 = 1/10+1/20.
R3 = 6.666666.

I =V/R = 24/26666.67 = 90 m A.

V = IR.
I = V/R.

R1 = R2 = R3 = 10 kohm.

One 20 kohm resistor is parallel with one of the 10 kohm resistor.

So say R4 = 20 kohm parallel with R1.

Total resistance= (R1*R4/R1+R4)+R2+R3 = 26666.66 ohms.

I = 24/26666.66 = 900 uA.

R=20+(20*10)/30=26.67.

I=V/R=24/26.67=.899 mA= 900uA.

Thanks nadarajan.

10K 10K 10K
------^^^^^-----^^^^-----^^^^------
| | |
| | |
| > |
| >20K OHM 24 V
| > |
| | |
------------------------------------ Is this circuit diagram is correct?

[{(20*10)/30}+10+10]=80/3 then
v=24v now
i=v/r=24/(80/3)*1/1000=(72/80)1/*1000=0.9/1000=0.0009=900microA

@Nadarajan :Thank you