Electrical Engineering - Series Circuits - Discussion

Discussion Forum : 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
69 mA
230 mA
23 mA
690 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Satheesh said:   9 years ago
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...

Vignesh kalyan said:   1 decade ago
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.

Pavithra said:   1 decade ago
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.

Anu said:   10 years ago
Hi @Vignesh, our aim is to find total current through the circuit for that we have to use the total resistance in account.

Navin said:   1 decade ago
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.
(1)

Ishwarya said:   1 decade ago
Thank you narender yadav.

Anand said:   1 decade ago
Nice explanation Narender

Prathisha said:   8 years ago
Thank you @Satheesh.

Narender yadav said:   1 decade ago
470/(3x680)=230 mv
(1)

Pavan nandigama said:   9 years ago
Thank you @Navin.


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