Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 35)
35.
What is the power dissipated by R1, R2, and R3?
Discussion:
7 comments Page 1 of 1.
Trxapeut said:
5 years ago
I think that the 0.52 W is wrong and that 0.46 W should be the answer.
So, it should be P1=0.26 P2=0.46 & P3=0.23.
So, it should be P1=0.26 P2=0.46 & P3=0.23.
(1)
Gabbie said:
6 years ago
I1 = 0.26 W ; I2 = 0.46 W; I3 = 0.92 W. These are my answers.
(1)
Peter. said:
9 years ago
I also calculated 0.46 W.
I also put this into circuit simulator, same result.
I also put this into circuit simulator, same result.
Dave said:
9 years ago
Yes, I got the answer for P2 is 0.46W.
Can someone explain the correct method?
Can someone explain the correct method?
Jess said:
9 years ago
How to solve the question?
My answer for P2 was 0.46W, am I doing something wrong?
My answer for P2 was 0.46W, am I doing something wrong?
Eduardo said:
1 decade ago
For the total resistance I got: 3666.67 Ohms.
For I1= 16.1 mA squared this & times by R1 --> PDr1= 258 mW.
For I2= 10.73 mA squared this & times by R2 --> PDr2= 460 mW.
For I3= 5.37 mA. squared this & times by R3 --> PDr3= 230 mW.
I am getting the right answers for P1 & P3. For P2, I am a little off. Am I doing something wrong?
Can someone explain me?
For I1= 16.1 mA squared this & times by R1 --> PDr1= 258 mW.
For I2= 10.73 mA squared this & times by R2 --> PDr2= 460 mW.
For I3= 5.37 mA. squared this & times by R3 --> PDr3= 230 mW.
I am getting the right answers for P1 & P3. For P2, I am a little off. Am I doing something wrong?
Can someone explain me?
Gopal said:
1 decade ago
Find out total resistace i.e. 11/3kohm
Find out current i.e. 59/(11/3)mA= 16.09mA
Now by formula i2R find out power dissipated
Find out current i.e. 59/(11/3)mA= 16.09mA
Now by formula i2R find out power dissipated
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