Electrical Engineering - Series Circuits - Discussion

Discussion Forum : Series Circuits - General Questions (Q.No. 4)
4.
A certain series circuit consists of a 1/8 W resistor, a 1/4 W resistor, and a 1/2 W resistor. The total resistance is 1200 . If each resistor is operating in the circuit at its maximum power dissipation, total current flow is
27 mA
2.7 mA
19 mA
190 mA
Answer: Option
Explanation:

Total P = P1 + P2 + P3 = 7/8w.

P = I2R.

I = sqrt (P/R)

Now, P = 7/8w, R = 1200 ohm.

Therefore,

I= sqrt ( (7/8)/1200 ) = 0.0270030862 = 27 mA.

Discussion:
32 comments Page 1 of 4.

Mitra rai said:   3 years ago
*I2 =(7/8)*(1/1200).
= √ of 0.000729.
= √ of 729*10^-6 = √729*(10^-3)2.
=27*10^-3.
=27 milli amp.

Charles Alex said:   3 years ago
Thanks for explaining the answer.
(2)

Rohine said:   3 years ago
I2 =(7/8)*(1/1200).
= √ of 0.000729.
= √ of 729*10^-6 =sqrt root of 729*(10^-3)2.
=27*10^-3.
=27 milli amp.

Pravali Bellana said:   3 years ago
Thank you for the explanation.

Parth Zala said:   4 years ago
Thanks everyone for giving the explanation.

Hayasi said:   5 years ago
Thanks all.

Yuvaraj said:   6 years ago
The current is same for all the components in a series circuit and here, all three resistors have different power ratings which means only one of the three resistor is operating at its maximum power dissipation and others are consuming less or more amount of power compared their power rating.

Subhash said:   7 years ago
How it come √(7/8)/1200= 0.027003?

Please describe.

Vipin said:   7 years ago
Please explain the last step of this answer.

Chaman chauhan said:   8 years ago
How it come sqrt of (7/8)/1200= 0.027003?

Please describe.


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