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
. If each resistor is operating in the circuit at its maximum power dissipation, total current flow isAnswer: 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 3 of 4.
Susanta said:
9 years ago
When power are series it calculate as equivalent capacitance.
Megharaj said:
9 years ago
How to calculate sqrt (7/8) /1200 without calculator? Anybody, please tell me.
Chaman chauhan said:
8 years ago
How it come sqrt of (7/8)/1200= 0.027003?
Please describe.
Please describe.
Vipin said:
7 years ago
Please explain the last step of this answer.
Subhash said:
7 years ago
How it come √(7/8)/1200= 0.027003?
Please describe.
Please describe.
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.
Hayasi said:
5 years ago
Thanks all.
Parth Zala said:
4 years ago
Thanks everyone for giving the explanation.
Pravali Bellana said:
4 years ago
Thank you for the explanation.
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.
= √ of 0.000729.
= √ of 729*10^-6 =sqrt root of 729*(10^-3)2.
=27*10^-3.
=27 milli amp.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers