# What is the fraction of current taken by this resistor?

A network of resistors has a pair of input terminals AB connected to a DC supply and a pair of output terminals CD connected to a load resistor of 120 Ohm. The resistances of the network are AC=DB=180 Ohm and AD=BC=80 Ohm.

What is the ratio of the current in the load resistor to that taken from the supply?

NB: Labels marked in red are for the mesh analysis I've made!

simulate this circuit – Schematic created using CircuitLab

The question is probably not very hard, however I am trying to find the answer by not applying Norton's or Thevenin's theorems, but only Kirchhof's laws and therefore Nodal and Mesh analysis.

What I've tried so far:

delta-star transformation With that it's easier to find the total resistance, but that's it.

simulate this circuit

nodal analysis ended up with (with reference to the initial diagram)

$$\frac{C-A}{180} - \frac{B-C}{80} = \frac{D-C}{120}$$ $$\frac{D-C}{120} = \frac{B-D}{180} - \frac{D-A}{80}$$

It seems be a deadlock with to many unknown values.

Finally mesh analysis, after solving simultaneous equations assuming current to flow clockwise in every closed loop (DC source connected in between points A and B) I've got following. $$I_{1} = \frac{V}{120}, I_{2} = \frac{V}{200}, I_{3} = \frac{V}{300}$$

UPD: Numerical answer for this question is (negative direction) 5 and 0.2 (for positive direction of current flowing), but the most important is how this answer was derived and ultimately this is my question.

The current through the load resistor can be obtained by finding the THEVENIN EQUIVALENT across it.If you are unaware about do check out thevenin intro or any other online material on it.It is very useful in solving networks for currents and voltages.

The thevenin equivalent across 120 ohm is :

From the above circuit the current through load can be easily calculated as:

I(Load) = [ 0.385Vs /(110.76 + 120) ] = 0.0016684 Vs ( where Vs is the supply voltage at input)

Now, to get the current drawn from the supply voltage Vs at the input,

Step 1: Find the equivalent resistance across the input source which involves star to delta conversion as you have mentioned.

The R(equivalent) = [ 37.9 + ( (56.8 + 80) || (25.3 + 180) ) ] = 120 ohms.

Step 2:

The current drawn from the input is : I(supply) = Vs/ Req = 0.00833 Vs.

Now that we have both currents in terms of the input supply voltage,

The ratio of load current to the current drawn from the supply is:

I(Load) / I(supply) = 0.001664 /0.00833 = 0.2

If the answer is 5 then it should be the inverse of the above as current through load is always less than supply current.

Edit: Since you wanted to solve it by mesh analysis here is an easier way:

• Thank you very much. I am fully aware of that theorem and I know how to apply it, however I was interested in finding the answer without using Thevenin's theorem. Or another question if that will bring us closer to the solution: If it is reasonable to solve that without Thevenin's theorem. If yes, then I am interested in that solution. – hrust Nov 17 '15 at 21:19
• I hope i have put up the solution you needed, do mark it as the solved to close this thread (click on the tick icon if you think the solution i posted is complete and helpful ). – Ashik Anuvar Nov 17 '15 at 21:55
• Thank you for the great job you done for me. As soon as I get my 15rep I'll upwote your answer, but it's the only think I can do now. – hrust Nov 17 '15 at 21:57
• If you are unaware of the matrix method, do look out for online material on the same or videos – Ashik Anuvar Nov 17 '15 at 21:59

Since I3 is the current flowing through the loop of the supply and the output resistor is common to loop 1 and 2, I1 - I2 will be the current flowing through it, i.e., the output current. So, I guess the answer will be the ratio of I3 and I1 - I2.

Edit

simulate this circuit – Schematic created using CircuitLab

Let the current in Loops 1, 2, 3 be I1, I2, I3 resp. The answer is independent of the source voltage, so assuming a value will make the solving part easier and won't change the answer. The Mesh matrix is, \begin{bmatrix} 380 & -120 & -80\\ -120 & 380 & -180\\ -80 & -180 & 260 \end{bmatrix} If the source voltage was 10V, the I1 = 0.0333, I2 = 0.05, I3 = 0.0833. So I3 / I2 - I1 = 4.989 which can be approximated to 5.
PS: You can try without assuming a value for source voltage

• You probably got mislead by the labels I've made. Those are for the mesh analysis I have performed. – hrust Nov 17 '15 at 20:45
• Ok, the power source goes across AB and it falls in the loop 3. The current in that loop will be the current drawn from the source. The output resistor lies in common to loop 1 and 2. So, the difference in the currents through these loops will be the output current. Did I get it correct this time? If yes, then their ratio will be your answer. – Raamakrishnan A. Nov 17 '15 at 20:52
• Using results from mesh analysis and substituting them in equation $$\frac{I_{3}}{I_{2}-I_{1}}$$ gave me -1 as a result. And result so far is not correct. – hrust Nov 17 '15 at 21:06
• Raamarkrishnan, stackexchange is answer-centric. If you have have an update to your answer, edit your answer above. Don't just leave it as a comment. – Dan Laks Nov 17 '15 at 21:07
• @Synaps check out my solution hope it helps – Ashik Anuvar Nov 17 '15 at 21:17