Maximum power transfer and kirchoffs

[Andrew187]

Hi Folks
I need a bit of help, My weakness is kirchhoffs law and algebra, I know but I still give it a try. For this task I am required to calculate the value of the load (RL) using kirchhoffs and then I have to find the maximum power transfer for the circuit. I have attempted to calculate the load (RL) using kirchhoff now the bit that I am confused on is am I looking for the resistance of the Load (RL) or the current in the load (RL). Which figure is required to calculate the maximum power transfer? is it the current or resistance of the load (RL)? If it is the current is my calculations correct? and if its the resistance I need to calculate, how do I figure that out because I have no clue where to start:(

 

[gneill]

The value that you found for I₂ is okay. Things go off the rails when you get to the section "Sub I₂ into equation (1). You write:
$$I_1 = 15v + \frac{0.293a}{15v} = 15.02$$
which doesn't make sense. The units are not consistent; current is not the same as voltage. Later you have RL = 15.03a, but RL is a resistance, not a current.

Can you state the "Max Power theorem"? Have you looked at Thevenin's theorem (and hence Thevenin equivalent circuits) yet?

 

[Andrew187]

This is my weakness this assignment is long overdue, I covered this last year now when I am going back through my notes it doesn't make sense:( But come to think of it yes you are correct I need to find the resistance of RL and then use that to find the maximum power transfer. I have covered thevenins and nortons but this task has to be calculated using kirchhoffs to find the resistance of RL, I am lost.

 

[gneill]

Use KVL and KCL to find the Thevenin equivalent. Then apply the Max Power theorem to find RL.

 

[Andrew187]

I still don't get it by thevanin equivalent and why I should use thevanin equivlent for this? I thought I had to use max power theorum once I determine the resistance for RL?

 

[gneill]

I still don't get it by thevanin equivalent and why I should use thevanin equivlent for this? I thought I had to use max power theorum once I determine the resistance for RL?

That's exactly backwards Find the Thevenin equivalent first, than add the load RL and use the Max Power theorem to find the value of RL that maximizes the power in RL.

 

[Andrew187]

I don't get it can't I find the resistance of the load (RL) by using Kirchoffs? and then find the maximum power transfer using the maximum power theorum?

 

[gneill]

I don't get it can't I find the resistance of the load (RL) by using Kirchoffs? and then find the maximum power transfer using the maximum power theorum?

Nope. Any value of RL will form a valid circuit that can be analyzed with KVL/KCL. There's no way to pick a particular value of RL from that.

The problem wants you to find the value of RL that maximizes the power transfer. To do that you need to apply the maximum power theorem. To apply the maximum power theorem you need to simplify the circuit to a form where it can be applied. The Thevenin equivalent is such a circuit.

So. First use Thevenin to reduce the network that will drive the load to its Thevenin equivalent. Then apply the max power theorem to the result to determine RL.

 

[Andrew187]

I'm lost now, it isn't making sense to me at all.

 

[gneill]

I'm lost now, it isn't making sense to me at all.

Perhaps you should review the Max Power Transfer theorem. Take a look at the circuit used to prove it; The Max Power theorem allows you to find the load resistance RL that yields the maximum power transfer from a given source voltage with a series resistance.

This is the same form you should reduce your circuit to in order to apply the theorem to find your value of RL.

 

[Andrew187]

So from what point do I change what I have done in my work? or do I just start all over again?

 

[gneill]

So from what point do I change what I have done in my work? or do I just start all over again?

You'll have to decide how you want to go about finding the Thevenin equivalent for the network. You may be able to salvage some of the work you've done. For example, your value of I2 will let you find the open circuit voltage at the network terminals.

 

[Andrew187]

OK I am going to salvage the first part which for I2 I got 0.293a I believe the second bit is nonsense so how do I calculate the voltage output (Vo=V_EF) without load (Rl=∞)?

 

[gneill]

OK I am going to salvage the first part which for I2 I got 0.293a I believe the second bit is nonsense so how do I calculate the voltage output (Vo=V_EF) without load (Rl=∞)?

What component is the Vo across?

 

[Andrew187]

What component is the Vo across?

Is it across all the resistors, R7,R8,R9,R10?

 

[gneill]

Is it across all the resistors, R7,R8,R9,R10?

Pick one... which resistor would you place the leads of a meter across in order to measure Vo?

 

[Andrew187]

Pick one... which resistor would you place the leads of a meter across in order to measure Vo?

Between R8 and R10 ?

 

[gneill]

Between R8 and R10 ?

No. The load (RL) is not connected between R8 and R10. What component is RL in parallel with?

 

[Andrew187]

R10 so would I measure across E and F?

 

[gneill]

R10 so would I measure across E and F?

Right.

Now, where does your I2 flow...

 

[Andrew187]

I2 flows in R8 R9 and R10?

 

[gneill]

I2 flows in R8 R9 and R10?

Yes, so what's Vo?

 

[Andrew187]

Yes, so what's Vo?

R8+R9+R10/ 0.293a= 25Ω+7Ω+24Ω/ 0.293a= 5.23v is that correct?

 

[gneill]

R8+R9+R10/ 0.293a= 25Ω+7Ω+24Ω/ 0.293a= 5.23v is that correct?

No. Vo is across R10 only! You know the current through R10. Vo is equal to the potential drop across R10. So use Ohm's law on R10.

 

[Andrew187]

No. Vo is across R10 only! You know the current through R10. Vo is equal to the potential drop across R10. So use Ohm's law on R10.

V =IR= V= 0.293a*24Ω=7.032v

 

[gneill]

V =IR= V= 0.293a*24Ω=7.032v

Yes. (If you keep a few more decimal places for intermediate values you'd find 7.037 V).

So that's the open circuit potential for the circuit, and thus the Thevenin voltage V_th.
What's the next step for finding the Thevenin equivalent?

 

[Andrew187]

I think its Isc? but for that I need to find I1 so how would I go about that?

 

[gneill]

You can either find Isc or directly determine the equivalent resistance of the network as seen from Vo.

To find Isc, redraw the circuit with the short in place and determine the current. To find Req, suppress the source and determine the equivalent resistance via simplification of the network.

 

[Andrew187]

Is the drawing correct?

 

[gneill]

Looks good.

 

[Andrew187]

Looks good.

Ok we are getting there lol I just need a clue how to work out I1 as I believe I need that for the ISC?

 

[gneill]

Ok we are getting there lol I just need a clue how to work out I1 as I believe I need that for the ISC?

Are you going to find the equivalent resistance from your diagram, or go the Isc route?

When determining the Thevenin equivalent you want to determine the Thevenin voltage Vth (the open circuit voltage, which you've done), and the Thevenin resistance. The Thevenin resistance can be found in two ways: 1) determine the short circuit current Isc for the network and divide Vth by Isc, or 2) suppress the sources and determine the equivalent resistance of the network looking back into it from the output.

In this case, I think the second method looks simpler as it's a pretty basic network with opportunities for simplification.

 

[Andrew187]

Isc=
(9+25)*i1 - 25*i2 = 15
-25*i1 + (25+7)*i2 = 0

 

[gneill]

Isc=
(9+25)*i1 - 25*i2 = 15
-25*i1 + (25+7)*i2 = 0

yes, and so,...

 

[Andrew187]

yes, and so,...

I2 = 0.81a?