Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 5)
5.
Find the current through R2 of the given circuit.
Discussion:
21 comments Page 2 of 3.
Vignesh said:
10 years ago
Please tell this problem.
Amke naveen said:
1 decade ago
Use super position theorem first consider dc voltage source open current source.
So i1 = 12/270+120.
Then current source i2 = 150m*270/270+120.
I = i1+i2.
So i1 = 12/270+120.
Then current source i2 = 150m*270/270+120.
I = i1+i2.
Pk pandey said:
1 decade ago
I make use of kcl but not able to get ans.
Omprakash jatav said:
1 decade ago
By using kvl, we can solve the easily final result 134.12.
Diksha said:
1 decade ago
By using nodal analysis, we can easily get the final result 134 mA.
Alwazzan said:
1 decade ago
I1 = I2 - 0.150 ---- 1.
v = 270I1 + 120I2 ---- 2.
12 = 270(I2-0.150) + 120I2
I2 = (12+270*0.150)/390 = 134 mA.
v = 270I1 + 120I2 ---- 2.
12 = 270(I2-0.150) + 120I2
I2 = (12+270*0.150)/390 = 134 mA.
Ruplal rajak said:
1 decade ago
Use of source transformation techniqu make will make the problem easy,120*0.15=18.00v will connected with series with 120 ohm resister and then apply kvl 120i+270i+18-12=-6/390
Archana said:
1 decade ago
0=V-12/270+V/120+150
V=16.15
current through R2
I=V/R
I=16.15/120
=134mA
V=16.15
current through R2
I=V/R
I=16.15/120
=134mA
M.V.KRISHNA/PALVONCHA said:
1 decade ago
Use nodal analysis, you will get nearer value to option D.
Mahesh kumar tanwar said:
1 decade ago
Why we can't apply kcl. I think kcl make it eassy.
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