Electrical Engineering - Series Circuits - Discussion

Discussion :: Series Circuits - General Questions (Q.No.12)

12. 

Three 680 resistors are connected in series with a 470 V source. Current in the circuit is

[A]. 69 mA
[B]. 230 mA
[C]. 23 mA
[D]. 690 mA

Answer: Option B

Explanation:

No answer description available for this question.

Narender Yadav said: (Feb 12, 2011)  
470/(3x680)=230 mv

Ishwarya said: (Mar 22, 2011)  
Thank you narender yadav.

Anand said: (Jul 26, 2011)  
Nice explanation Narender

Pavithra said: (Jul 23, 2012)  
The resistors are connected in series. Therefore the total resistance is 680+680+680=2040.

By ohms law, V=IR.

I=V/R.

=470/2040.

=230 milli-ohms.

Navin said: (Dec 15, 2013)  
I = V/R, BY ohm's law,

R = R1+R2+R3.
= 680+680+680.

= 2040, V=470V.

I = 470/2040.
= 0.2303921A*10^3 = 230mA.

Vignesh Kalyan said: (Mar 2, 2015)  
My doubt is that in the series circuit current through each resistor is same, then why we cannot find the individual current for single resistor.

v = 470 v, R = 680 ohm, then i = v/r.

i = 470/680,

i = 69 ma,

So the current is same, the answer is 69.

Anu said: (Nov 3, 2015)  
Hi @Vignesh, our aim is to find total current through the circuit for that we have to use the total resistance in account.

Satheesh said: (Nov 25, 2016)  
Hello @Vignesh Kalyan.

Thanks for your doubt that many will have the same. Let me clarify your query.
As per ohm's law current is the same throughout the series circuit but voltage dividers, we can't consider 470 V for single resistors.

I hope you will understand...

Pavan Nandigama said: (Feb 14, 2017)  
Thank you @Navin.

Prathisha said: (Mar 22, 2017)  
Thank you @Satheesh.

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