Electrical Engineering - Circuit Theorems and Conversions - Discussion

Discussion :: Circuit Theorems and Conversions - General Questions (Q.No.5)

5. 

Find the current through R2 of the given circuit.

[A]. 30.7 mA
[B]. 104 mA
[C]. 74 mA
[D]. 134 mA

Answer: Option D

Explanation:

No answer description available for this question.

Pk Pandey said: (Feb 8, 2011)  
I make use of kcl but not able to get ans.

Deepa said: (Jul 15, 2011)  
Do oc of current and apply ohms law so current=31ma
and sc voltage source apply current divider rule then you obtain current=103.84ma
add both you get 134ma(approx).

Mahesh Kumar Tanwar said: (Nov 9, 2011)  
Why we can't apply kcl. I think kcl make it eassy.

M.V.Krishna/Palvoncha said: (Dec 18, 2011)  
Use nodal analysis, you will get nearer value to option D.

Mohammad Saddam said: (Jun 7, 2012)  
Using superposition theorem.

Archana said: (Aug 27, 2012)  
0=V-12/270+V/120+150
V=16.15
current through R2
I=V/R
I=16.15/120
=134mA

Ruplal Rajak said: (Sep 4, 2012)  
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

Alwazzan said: (Jul 24, 2013)  
I1 = I2 - 0.150 ---- 1.

v = 270I1 + 120I2 ---- 2.

12 = 270(I2-0.150) + 120I2

I2 = (12+270*0.150)/390 = 134 mA.

Diksha said: (Nov 21, 2013)  
By using nodal analysis, we can easily get the final result 134 mA.

Omprakash Jatav said: (Jan 27, 2014)  
By using kvl, we can solve the easily final result 134.12.

Sandeep said: (Feb 6, 2014)  
By applying nodal analysis, instead of -150 why +150 is represented.

Amke Naveen said: (Dec 5, 2014)  
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.

Dusmanta Rohiodas said: (Apr 13, 2015)  
0 = V-12/270 + V/120 + 150.
V = 16.15.

Current through R2.

I = V/R.
= 16.15/120.
= 134 mA.

Vignesh said: (Nov 9, 2015)  
Please tell this problem.

Jagadesh said: (Feb 23, 2016)  
By applying nodal analysis.

Current entering = Current leaving.

Assume the 120 ohm branch node as v.

150 mA = ((V - 12)/270) + (V/120).

0.15 = (13 V - 48)/1080.

162 = 13V - 48.

13 V = 210.

V = 210/13 V.

Current in 120 ohm branch = V/120.

= 210/(13*120).

= 7/(13*4) = 7/52 = 0.134.

= 134 mA.

Basireddy said: (Feb 9, 2017)  
Write loop equation.

270(i1)+120(i1+i2)=12V ..........eqn-1
And i2=150mA.

Now solving eqn-1.
270(i1)+120(i1)+(120*150mA) = 12V
390(i1)+18V = 12V
390(i1) =12V-18V
390(i1) = -6V
i1 = -6V/390
i1 = -15.38mA

Now current flowing through R2 is (i1 + i2).
Therefore, -15.38mA + 150mA = 134.62mA.

D.Mahammad Ali said: (Mar 13, 2017)  
By nodal annalysis first find node voltage then find the current.

Md, Rayhan Uddin said: (Jul 12, 2017)  
0 = V-12/270 + V/120 " 0.15,
=>(4V-48+9V-162)=0,
=>13V=210,
=>V=210/13,
=>V=16.15V,
Since, IR2=16.15/120,
=134mA.

Fayaz said: (Dec 9, 2017)  
16.15/120=.134amp.
134/1000=1.34ma.

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