Electrical Engineering - Series-Parallel Circuits - Discussion

Discussion :: Series-Parallel Circuits - General Questions (Q.No.6)

6. 

Three 10 k resistors are connected in series. A 20 k resistor is connected in parallel across one of the 10 k resistors. The voltage source is 24 V. The total current in the circuit is

[A]. 900 A
[B]. 9 mA
[C]. 90 mA
[D]. 800 A

Answer: Option A

Explanation:

No answer description available for this question.

Ramu said: (Jun 10, 2011)  
(30+20+10)/0.066 = 900

Nadarajan V said: (Jun 22, 2011)  
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.

Anand said: (Jul 27, 2011)  
Nice Explanation Nadarajan...

Pradeep said: (Sep 17, 2011)  
@Nadarajan :Thank you

Uday said: (Sep 23, 2011)  
[{(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

Shailaja said: (Jan 16, 2012)  
10K 10K 10K
------^^^^^-----^^^^-----^^^^------
| | |
| | |
| > |
| >20K OHM 24 V
| > |
| | |
------------------------------------ Is this circuit diagram is correct?

Pukhraj said: (Mar 17, 2012)  
Thanks nadarajan.

Arun said: (Dec 12, 2012)  
R=20+(20*10)/30=26.67.

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

Trupti said: (Jun 23, 2015)  
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.

Ali said: (Sep 10, 2015)  
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.

Aamir said: (Nov 20, 2015)  
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.

Pappish said: (Feb 4, 2016)  
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.

Mayur Shankariya said: (Jun 14, 2016)  
@Uday.

Thanks for the explanation.

Shashikant Chiranjivi said: (Feb 7, 2017)  
Thanks @Arun.

Sourabh said: (Feb 7, 2017)  
Nice explanations, thank you all.

Sanjay said: (Mar 3, 2017)  
Thanks for explanation @Trupti.

Rion said: (Mar 29, 2017)  
Here,

R = 20k||10k + 20k.
I = 24/R.

Thiyanesh said: (Jul 9, 2017)  
Thanks @Ali.

Rakesh said: (Feb 22, 2018)  
How to solve it? please explain me.

Anomii said: (Aug 14, 2018)  
Good explanation. Thanks.

Coolman said: (Oct 17, 2019)  
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.

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