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 3 of 5.
Sankeerthana said:
7 years ago
Why the answer is different when we apply v^2/R?
Hanis said:
7 years ago
When you do the loop?
how do you get Equation 1: 11kI1-8kI2-111 = 0, where does the 11KI1 come from?
how do you get Equation 1: 11kI1-8kI2-111 = 0, where does the 11KI1 come from?
Ganesh said:
6 years ago
For individual resistance, I am getting R2 = 0.01773ohms.
Durga Prasad said:
4 years ago
Actually mesh analysis is better to get currents and then power.
But it takes so much time.
We have to solve this in the simplest way can anyone tell how to solve this?
But it takes so much time.
We have to solve this in the simplest way can anyone tell how to solve this?
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.
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
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers