Discussion :: Circuit Theorems and Conversions  General Questions (Q.No.5)
5.  Find the current through R_{2} of the given circuit.


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=V12/270+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+1812=6/390 
Alwazzan said: (Jul 24, 2013)  
I1 = I2  0.150  1. v = 270I1 + 120I2  2. 12 = 270(I20.150) + 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 = V12/270 + 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 ..........eqn1 And i2=150mA. Now solving eqn1. 270(i1)+120(i1)+(120*150mA) = 12V 390(i1)+18V = 12V 390(i1) =12V18V 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 = V12/270 + V/120 " 0.15, =>(4V48+9V162)=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. 
Anon said: (Dec 19, 2019)  
By nodal analysis we will arrived to this equation; (V12)/270 + V/120 150mA =0 . V=16.15V. Hence, I2=16.15/120. I2=1.346mA. 
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