Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 2)
2.
What is the power dissipated by R2, R4, and R6?
Discussion:
45 comments Page 2 of 5.
Mayank said:
8 years ago
Node connecting r1, r2, r3-X say X volt also,
.............................. r3, r4, (r5+r6) - Y.
Apply nodal analysis.
(x-111)/3+(x-y) /2+(x/8)=0.
(y-x)/2+y/7+y/(1+6)=0,
Solve X and Y.
Use power=V*V/R.
For r2 p2=(x-y) ^2/2,
P4=y^2/7,
Don't have to solve p6.
.............................. r3, r4, (r5+r6) - Y.
Apply nodal analysis.
(x-111)/3+(x-y) /2+(x/8)=0.
(y-x)/2+y/7+y/(1+6)=0,
Solve X and Y.
Use power=V*V/R.
For r2 p2=(x-y) ^2/2,
P4=y^2/7,
Don't have to solve p6.
Baby said:
8 years ago
I agree @Arun.
D. Mk said:
8 years ago
Easy to follow your solution. Thanks @Arun.
Sana said:
8 years ago
How do we get the loops? Describe it in detail.
EMEDIONG said:
9 years ago
Please, can someone please take it step by step for me to follow.
Saipriya said:
9 years ago
Please tell, how to get current in each resistor?
(1)
Nagesh said:
9 years ago
By using superposition theorem.
We will find out current at resistors, so by using i*i*r we find power.
We will find out current at resistors, so by using i*i*r we find power.
(1)
Sue said:
9 years ago
By using the formula p = I^2/R.
WE CAN CALCULATE POWER IN R2, R4 AND R6.
WE CAN CALCULATE POWER IN R2, R4 AND R6.
Aseel said:
9 years ago
Yes, I am also not able to understand this, please repeat it and make it clearer.
Richard Msesi said:
9 years ago
I don't understand guys please can you put it in a simple way? i.e. use ohm's law.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers