Electrical Engineering - Series-Parallel Circuits - Discussion

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

Ohms Law I = V/R.

So, As question said 3 resistors in series with each value 10K and second question said 20K resistor parallel connected with one of 10K resistor.

Therefore, we can say total resistance is = R1+R2+(R3//20K), hence R3 resistor parallel connected with 20K,

= 10K + 10K+((10K*20K) / (10K+20K).
= 26.67K ohm.

So now the total current is,
I = V/R.
= 24V / 26.67K.
= 900 micro ampere.

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.

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.

across one of the resistors 20 ohms placed in parallel .so solving parallel formula
R1*R2/R1+R2=>20*10/20+10=6.66 .

Other two resisters with parallel solved value =6.66+10+10=26.66.

I=V/R => 24/26.66=900*10^-6.

20k ohm + ((20*10)/(20+10)) = 26.67k ohm

V= IR => I= V/R
=> I= 24V/26.67k ohm
=> I= 24V/26670ohm
=> I= 0.0008998875A
=> I= 899.8875 micro Amps
so 900 micro Amps is the correct answer.

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.

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.

[{(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

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

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