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.
Aseel said:
9 years ago
Yes, I am also not able to understand this, please repeat it and make it clearer.
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.
EMEDIONG said:
9 years ago
Please, can someone please take it step by step for me to follow.
Sana said:
8 years ago
How do we get the loops? Describe it in detail.
D. Mk said:
8 years ago
Easy to follow your solution. Thanks @Arun.
Baby said:
8 years ago
I agree @Arun.
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.
Aarti said:
8 years ago
Very nice answer and question because answer.
Abhinav said:
8 years ago
Can we not apply p = v^2/R?
Gopala R said:
7 years ago
Piease explain clearly.
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