Electrical Engineering - Circuit Theorems and Conversions - Discussion

Discussion :: Circuit Theorems and Conversions - General Questions (Q.No.3)

3. 

Find the total current through R3 in the given circuit.

[A]. 7.3 mA
[B]. 5.5 mA
[C]. 12.8 mA
[D]. 1.8 mA

Answer: Option D

Explanation:

No answer description available for this question.

Beth Tate said: (Oct 28, 2011)  
If you do the node analysis on this problem, you get your current to be .769 mA. By splitting the above circuit into two sections with 2 diffferent currents, you also get .769 mA. How do you come to get the answer 1.8 mA ?

Krishnan said: (Nov 21, 2011)  
Can any one explain the answer more clearly?

M.V.Krishna/Palvoncha said: (Dec 18, 2011)  
You are right Beth Tate.

Ronit said: (Feb 11, 2012)  
The answer I'm getting is 0.77mA.

Can anyone explain it - How do they get the answer 1.8mA ?

Reddi said: (Mar 1, 2012)  
We apply the kirchoff current law
(V-24/1.2)+(V+18/1.2)+(V/3.3)=0
we get V=2.55Mv
I=V/R=(2.55/3.3)=0.77mA

Adnan Cecosian said: (May 2, 2012)  
Use mesh analysis:
Create two loops then apply KVL we get 2 equations

Loop1; (1.2k)(I1)+(1.2k)(I1+I2)=24
Loop2: (1.2k)(I2)+(1.2k)(I1+I2)=-18
Rearrange these two equations
then find Whole Deterimnent which i found
=4.32k^2
use cramer Rule
I1=18.35mA
and I2=16.6mA
now therefore current through 1.2kohm Resistor is
I1+I2=18.35-16.6
=1.75mA
So Answer D is correct.

Bettycroger said: (Aug 22, 2012)  
The question asked for the current through R3 = 3.3kohm

Wee Kian said: (Sep 1, 2012)  
So is R2 or R3. the diagram is so misleading. does the VS1 connected to ground or just R3?

Pocahontas said: (Nov 3, 2013)  
The answer that I've got is 0.769mA using the superposition theorem.

Harsh Sharma said: (Feb 7, 2014)  
If you want the answer by nodal analysis,

See that the 18 we source is coming in front of 24 we source because in between their path no element is connected.

So while writing kcl we have [ 24 - 18 ] = 6 we source above the 1.2k resistor.

By that you solve and will get the answer as 1.8 mA.

Answer can also be given by source transformation technique.

Try it once also interesting to analyse the circuit.

Mayur said: (Apr 21, 2014)  
24-18 = 6v.
Total current through r3.
r3 = 3.3.

I = V/R.
I = 6/3.3.
I = 1.80.

Bhushan R Chaudhari said: (Sep 20, 2014)  
Mayur you are right, common point is 6 volt higher than ref. point i.e. voltage across 3.3kohm resister is 6v. therefore current through it is 6v/3.3kohm = 1.8mA.

Sfgsgsdgfdg said: (Oct 25, 2014)  
Oh, everyone R3 is not connected to ground since it does not have a dot. Instead R3 is connected to VS1.

Jan said: (May 6, 2015)  
Hi every one,

R2 and voltage source are connect ground potential.

Then I = V/R.

= 24/1.2+3.3 = 5.33 mA.

Why do you consider 18 volt source? That is open.

Please explain any one. Thanks advance.

Sathish said: (Jun 26, 2015)  
If we consider the first loop there are two voltage sources then why do we consider only 24v source alone? Can anybody answer my question?

Mayur said: (Aug 17, 2015)  
I used the Thevenin voltage theorem across 3.3 K ohm.

Vth = 3V and Zth = 600 ohm so by this I got answer as 0.78 mA.

Karthick S said: (Oct 3, 2015)  
What is a formula to find out the short circuit current in a transformer?

Jagadesh said: (Feb 23, 2016)  
As both the voltage sources are opposing in connection therefore the resultant voltage will be 24-18 = 6V.

The current through 3.3k resistor will be 6/3.3k = 1.8 mA.

Prabha said: (May 31, 2016)  
Superposition Theorem says:-

The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately.

To evaluate the separate currents to be combined, replace all other voltage sources by short circuits and all other current sources by open circuits.

By using this the answer is 5.5mA.

A Patki said: (Jul 23, 2016)  
I didn't understand the solution, please explain.

Sanju said: (Sep 29, 2016)  
Let's consider a point of junction of r2 and r1 has a potential V2.
Then by KCL: ((24 - V2)/r1) = ((V2 - (-18))/r2) + ((V2 - (ground voltage 0V) )/ r3)

We get V2 = 3/38 volts,
and I through register is V2/r3 = 7.3 mA.

Sanju said: (Sep 30, 2016)  
The actual answer is 0.7 mA. I run in the simulator and found that.

Basavaraj S Kumbalavati said: (Oct 27, 2016)  
From nodal analysis,

Node voltage is 2.538v.
Current through 3.3Kohm is (2.358/3.3) * mA = .769mA.

Swallah Marani said: (Nov 15, 2016)  
Using superposition.

Current produced by 24V = 11.5A
Current produced by 18V = 8.7A
Resultant current ; 11.5- 8.7 =2.8
Current in R3 3.3 ; 2.8[1.2]/[1.2+3.3] = 0.75A#.

Shivanya said: (Dec 17, 2016)  
By using source transformation we get .769 as an answer.

Raghukumar B S said: (Dec 30, 2016)  
24 - 18 = 6v.
Total current through r3.
r3 = 3.3k ohm.

I = V/R.
I = 6/3.3k ohm.
I = 1.80mA.

Rituraj Patra said: (Aug 12, 2017)  
The voltage across r3 is24-18 = 6, and current is 6/3.3=1.8.

Dev said: (Sep 21, 2017)  
I got 0.769 mA through 3.3 ohms resistor using KVL, node analysis, superposition according to my view answer is .769 mAmp.

Thanks.

Stoyan said: (Dec 20, 2017)  
The correct answer is 0.769 mA. I also agree.

Satyajit said: (Jan 12, 2018)  
The voltage across the 3300-ohm resistor is 2.538 v, not 6 volt due to drop in 1200 ohm resistors.
By nodal analysis, superposition theorem, Thevenin's theorem I got the answer 0.000769amp

Kiruthika said: (Jan 23, 2018)  
I also got the answer 0.769mA.

Bibek said: (Apr 21, 2018)  
@Mayur.

Can you help me by explaining how you have calculated Thevenin Voltage?

Sikandar said: (May 4, 2019)  
No, the Correct ans is 5.3mA. So option B is correct.

Sivakumar.G said: (May 28, 2019)  
24-18 = 6v.

Total current through r3.
r3 = 3.3.

I = V/R.
I = 6/3.3.
I = 1.80.

Anon said: (Dec 19, 2019)  
I don't agree with the 6V Vth since the circuit has aiding voltages.

Calculuseuler said: (Jun 29, 2020)  
The 1.2k and the 18v source, are open leaving 24v source connected in series with the other two.

So, the current through 3.3k.
= 24/3.3,
=7.3mA.

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