Solving problems using Tellegen's theorem. - Page 1
Forum section: electronics / beginners

author: avogadro
author_rank: Contributor
timestamp: March 11, 2019, 07:24:19 pm
content: Ive been trying to figure out this problem but I couldnt find solutions anywhere online. Ill attach the circuit. You have to find current through R5 when the switch is open and you only know values of R1, R5 and currents i1 and iR5 when the switch is closed. I have the solutions but I dont fully understand them and neither did the professor do good job at explaining it. I get the general gist of the theorem but in the solutions he's saying that the sum of the products of currents when switch is closed and voltages when the switch is open are equal to zero which I dont understand why.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2019, 07:53:31 pm
content: The theorem simply states that the power dissipated in all the resistors is equal to the power sourced by the battery.In other words:$$Ei_1=i_{R1}^2 R_1+i_{R2}^2 R_2+i_{R3}^2 R_3+i_{R4}^2 R_4+i_{R5}^2 R_5+i_{R6}^2 R_6$$However, it is not immediately obvious to me that there is sufficient information given to solve the problem. Are the values of the resistors known apart from \$R_1\$ and \$R_5\$?Maybe there is a trick to finding the solution without knowing all the values, but I would have to spend time working through it.

author: avogadro
author_rank: Contributor
timestamp: March 11, 2019, 08:20:11 pm
content: There is a trick. Ill attach the solution. I dont really get how I can use this and why this works. He just redraws the circuit and pulls ratio i1/i1'=iR5/iR5' out of thin air.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2019, 09:38:33 pm
content: I am having difficulty seeing why equation 1 is true.What I would do as a student is plug some actual numbers into the problem for the voltage E and the various resistances and calculate the currents from first principles with the switch open and closed. Does the equation given as the solution hold true for various numerical examples? That would at least convince me that the theory is correct, even though I don't understand it.As to the worked solution, it needs someone else to help as I would have to think about it a lot more to understand why the derivation holds true.

author: Wimberleytech
author_rank: Super Contributor
timestamp: March 12, 2019, 01:13:04 am
content: Starting at equation 2 is straightforward. Eq 2 simply states that the equivalent resistance looking into the network at R2 does not vary based on the switch being open or closed.The derivation from eq 2 to eq 3 is straightforward.The next trick is recognizing that i5 = K i1 and K is linear time invariant. Then the ratio of i1 before and after gives the ratio of i5 before and after where K cancels out.I agree that equation 1 and taking the voltage across a short seems odd.

author: IanB
author_rank: Super Contributor
timestamp: March 12, 2019, 02:26:40 am
content: Let me try from first principles:From linearity, we can observe that:$$\frac{\bar{u_1}}{u_1}=\frac{\bar{i_1}}{i_1}=\frac{\bar{i_{R5}}}{i_{R5}}$$It follows that:$$\bar{i_{R5}}=i_{R5}\frac{\bar{u_1}}{u_1}$$Also:$$\begin{align}u_1 & =E\\\bar{u_1} & =E-\bar{i_1}R_1\end{align}$$From which:$$\frac{E-\bar{i_1}R_1}{E}=\frac{\bar{i_1}}{i_1}$$Solving for \$\bar{i_1}\$:$$\bar{i_1}=\frac{E i_1}{E+i_1 R_1}$$Therefore:$$\bar{i_{R5}}=i_{R5}\frac{\bar{u_1}}{u_1}=i_{R5}\frac{\bar{i_1}}{i_1}=i_{R5}\frac{E}{E+i_1 R_1}$$Which is the result expected.

author: kulky64
author_rank: Regular Contributor
timestamp: March 12, 2019, 09:40:06 pm
content: OP stated that the known quantities are R1, R5 and i1, iR5 when the switch is closed. So I expected the result in the form iR5'=f(R1, R5, i1, iR5). I thought E is unknown.

author: IanB
author_rank: Super Contributor
timestamp: March 13, 2019, 12:47:03 am
content: Quote from: kulky64 on March 12, 2019, 09:40:06 pmGood point. But the given worked solution is in terms of E. Can it be solved if E is unknown?