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 4 of 5.
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.
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.
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)
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.
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.
Arun said:
1 decade ago
By mesh analysis inspection method:
11 I1 - 8 I2 = 111 --> EQN 1.
-8 I1 + 17 I2 - 7 I3= 0 --> EQN 2.
-7 I2 + 14 I3 = 0 --> EQN 3.
BY RESOLVING THESE EQUATIONS WE GET,
I1 = 17.72 mA
I2 = 10.50 mA
I3 = 5.25 mA
Current through R2 is I1-I2 = 7.22 mA.
And by P = (I^2)R formula we can calculate power across the resistor R2
P = 7.22*7.22*10^-6*8*10^3 = 417 mW.
11 I1 - 8 I2 = 111 --> EQN 1.
-8 I1 + 17 I2 - 7 I3= 0 --> EQN 2.
-7 I2 + 14 I3 = 0 --> EQN 3.
BY RESOLVING THESE EQUATIONS WE GET,
I1 = 17.72 mA
I2 = 10.50 mA
I3 = 5.25 mA
Current through R2 is I1-I2 = 7.22 mA.
And by P = (I^2)R formula we can calculate power across the resistor R2
P = 7.22*7.22*10^-6*8*10^3 = 417 mW.
Jorge said:
1 decade ago
Hi does anyone can tell me this example step by step?
Jorge said:
1 decade ago
Hello if is possible I would like to see the answer procedure step by step of this exercise. Thank you.
2. What is the power dissipated by R2, R4, and R6?
2. What is the power dissipated by R2, R4, and R6?
Yeyo said:
1 decade ago
More description anybody?
(1)
SONU PRADHAN said:
1 decade ago
Apply kvl to the loop and find current
Which is,
I1=0.01778
I2=0.01058
I3=0.00529
Now current across R2=0.0072
Now current across R4=0.00529
Now current across R6=0.00529
We Know that P=I2R
SO P across R2=(0.0072)2*R2
=(0.0072)2*8000
=417mW.
P across R4=193
P across R6=163
Which is,
I1=0.01778
I2=0.01058
I3=0.00529
Now current across R2=0.0072
Now current across R4=0.00529
Now current across R6=0.00529
We Know that P=I2R
SO P across R2=(0.0072)2*R2
=(0.0072)2*8000
=417mW.
P across R4=193
P across R6=163
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