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 1 of 5.
Hemanth said:
1 decade ago
Can you give description anybody?
Ranjan Singh said:
1 decade ago
First determine the current as I1,I2 and I3.
I1=0.01778A
I2=0.01058A
I3=0.00529A
Now current across R2=0.0072A
" " " R4=0.00529A
" " " R6=0.00529A
By using the formula for power as P=I2R
We will get power across R2=414mW
"""""""""""""""""""""""""R4=196mW
"""""""""""""""""""""""""R6=168mW
So answer will be [A].
I1=0.01778A
I2=0.01058A
I3=0.00529A
Now current across R2=0.0072A
" " " R4=0.00529A
" " " R6=0.00529A
By using the formula for power as P=I2R
We will get power across R2=414mW
"""""""""""""""""""""""""R4=196mW
"""""""""""""""""""""""""R6=168mW
So answer will be [A].
Avi said:
1 decade ago
How to determine current in this circuit. Will any body explain ?
Priya said:
1 decade ago
In this we use mesh analysis method for current flowing. where have to find
I1=delta1/delta;
I2=delta2/delta;
I3=delta3/delta
I1=delta1/delta;
I2=delta2/delta;
I3=delta3/delta
Ron said:
1 decade ago
Redraw the circuit to it's simplest series form by getting the total resistance and current for the entire circuit then work your way backwards by applying ohm's law to determined the other values.
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
Yeyo said:
1 decade ago
More description anybody?
(1)
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?
Jorge said:
1 decade ago
Hi does anyone can tell me this example 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.
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