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.
Vinay said:
1 decade ago
Can anybody please tell me what is total current in the circuit and the voltage drop at R2 step by step.
MeghA said:
1 decade ago
Since the voltage drop is same in parallel circuits why can't we take.
P = V^2/R formula.
P = V^2/R formula.
Abdul Haleem said:
1 decade ago
Using KVL on three loops, your equations will be:
Equation 1: 11kI1-8kI2-111 = 0.
Equation 2: -8kI1+17kI2-7kI3 = 0.
Equation 3: -7kI2+14kI3 = 0.
Solving the equations will give.
I1 = 17.73mA, I2 = 10.5mA, I3 = 5.25mA.
Go back to circuit you will find that,
P2 = (I1-I2)^2*R2 = 417mW.
P4 = (I2-I3)^2*R4 = 193mW.
P6 = I3^2*R6 = 166mW.
Equation 1: 11kI1-8kI2-111 = 0.
Equation 2: -8kI1+17kI2-7kI3 = 0.
Equation 3: -7kI2+14kI3 = 0.
Solving the equations will give.
I1 = 17.73mA, I2 = 10.5mA, I3 = 5.25mA.
Go back to circuit you will find that,
P2 = (I1-I2)^2*R2 = 417mW.
P4 = (I2-I3)^2*R4 = 193mW.
P6 = I3^2*R6 = 166mW.
(1)
Jehangir khan khaksar said:
1 decade ago
Make the circuit simple first such away that all will be in parallel and now cdr only apply for parallel those which will be in series are not consider.
First r5 and r6 make series this make big resistance parallel with r4 so new resistance is r456 this is in series with in r3 so r3456 which is parallel with r2 now apply but omit r1.
First r5 and r6 make series this make big resistance parallel with r4 so new resistance is r456 this is in series with in r3 so r3456 which is parallel with r2 now apply but omit r1.
Mahesh Reddy said:
1 decade ago
We can use this also I think. Please correct me if I am wrong.
P=V2/R. We know that voltage will be same across all the branches i.e. 111.
P=V2/R. We know that voltage will be same across all the branches i.e. 111.
Kim said:
10 years ago
Guys can somebody explain this step by step like finding required, voltage drops, currents of each resistors and also the power of each things.
Kim said:
10 years ago
Can somebody explain this step by step with units? So hard to understand this. Sorry guys. With total current, voltage drops, current and power for each resistors.
REETIKA said:
10 years ago
Please solve this circuit by using ohm's law.
Lara said:
10 years ago
I didn't know how did you find the power for each one?
Jorda said:
10 years ago
Pr2 = Vr2(I1-I2).
I1 = 0.01778 A.
I2 = 0.01058 A.
I3 = 0.00529 A.
= 57.9 V (0.01778 A-0.01058 A).
= 0.417 W.
Pr4 = Vr4 (I2-I3).
Pr6 = Vr6 (I3).
I1 = 0.01778 A.
I2 = 0.01058 A.
I3 = 0.00529 A.
= 57.9 V (0.01778 A-0.01058 A).
= 0.417 W.
Pr4 = Vr4 (I2-I3).
Pr6 = Vr6 (I3).
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