Where to attach the leads for the equivalent resistor?

[dimpledur]

So for each of the figures I want to find the Thevinin equivalent resistance and voltage. I also want to find the maximum power dissipated by a load resistor placed between points a and b on all three figures. Finally, I want to know what Vab is when a 7-olm resistor is placed between a and b. Thanks in advance for the help!

So for figure i) I found Rth=[(1/5)+(1/5)]^{-1}Ω+10Ω=12.5Ω
I found Eth=(5Ω)(5V)(5Ω+3Ω+2Ω)^{-1}=2.5V and hence I have the Thevinin equivalent circuit.

The maximum power dissipated is Eth^2/(4*Rth)=(2.5V)^2/(4*12.5Ω)=0.125 W.

And if there was a 7Ω resistor placed between a and b Vab=(7Ω)(2.5V)/(7Ω+12.5Ω)=0.897V

Next, for figure ii)
If you short circuit both of the voltage sources, you can move the 10Ω resitor to the side where the 3V source used to be, and hence you have 4 resistors in parallel. Thus,

Rth=[(1/(2Ω+3Ω))+(1/5Ω)+(1/10Ω)+(1/5Ω)]^{-1}=10/7Ω

Next, to find the Eth, we short circuit one source at a time and find the contribution of each source.
Due to the 3V source, we have:
Req=[(1/5Ω)+(1/12.5Ω)]^{-1}=3.57Ω

Thus, I=3V/3.57Ω=0.84A and therefore Eth=(0.84A)(5Ω)=4.2 V.

Next, due to 5 V source:

You can reduce the circuit to a single series circuit and Eth=(2Ω)(5V)/(2Ω+2Ω+3Ω)=10/7 V

and therefore, due to both sources Eth=10/7V+4.2V=5.63V. Hence, we have our Thevinin equivalent circuit.

The max power dissipated is =(5.63V)^2/(4*10/7Ω)=5.54 W

And if a 7Ω resistor was placed between ab we have Vab=(7Ω)(5.63V)/(10/7Ω+7Ω)=4.68V

Next, for figure iii)
When you short circuit the voltage supplys, you end up with 3 parallel resistances.

Rth=[(1/(2Ω+3Ω))+(1/5Ω)+(1/(10Ω+12Ω))]^{-1}=2.245Ω

Next, the Eth due to the 2V source:
Eth=(12Ω)(2V)/(12Ω+10Ω+2.5Ω)=0.9796V

Now, due to 5V source:
Eth=(4.074Ω)(5V)/(4.074Ω+3Ω+2Ω)=2.245V

and together Eth=2.245V+0.9796V=3.22V and we have our THevinin equivalent.

Pmax=(3.22V)^2/(4*2.245Ω)=1.16W

and Vab due to 7Ω resistor between ab is =(7Ω)(3.22V)/(2.245Ω+7Ω)=2.44V

 

[dimpledur]

Essentially I just want to know if I did the equivalents correct. I have a midterm this week and I have been struggling with this course, help is greatly appreciated!

 

[gneill]

(i) looks fine.

In (ii), the 3V source pins the output voltage at ab to 3V. Period.
If you short both supplies, the shorted 3V supply puts a short across ab, so the Thevenin resistance is zero.

in (iii), you don't end up with three parallel resistances when you short the supplies. The 12 Ohm resistor is across ab, but the 10 ohm resistor is in series with the parallel combination of the 5Ω with the (3+2)Ω resistances. It would be expedient in this case to recognize that the circuit from part (i) recurs in the left half of this circuit, adding just the 2V supply and the new 12Ω resistor. So just drop the Thevenin equivalent from (i) in its place of the recurring bit. Then simplifying the rest should be easy.

 

[dimpledur]

Thanks. I have a couple quick questions. I've noticed sometimes you can move around resistors, but in figure two you can't move the one resistor in the 3V sources place when its short circuited?
Lastly, what exactly determines which portions of a circuit are affected by a short circuit?

 

[gneill]

Thanks. I have a couple quick questions. I've noticed sometimes you can move around resistors, but in figure two you can't move the one resistor in the 3V sources place when its short circuited?
Lastly, what exactly determines which portions of a circuit are affected by a short circuit?

A short circuit joins nodes together and makes the potential difference zero. It cannot be anything else but zero. Nothing "behind" the short that has both its "legs" connected to that single node can affect the voltage at that node or net current for that node. So in the case of figure (ii), shorting the batteries yields the following:

 

[dimpledur]

So essentially a wire will cause a short circuit when that wires' nodes are in contact with BOTH nodes of a network? I don't really see how how the 10, 2, and 3-olm resistors are affected by it other than the fact that the voltage supplies are affected. Both nodes of those resistors don't appear to be in contact with both nodes of the wire causing a short.

 

[gneill]

Replace the resistor network to the left of the short with its equivalent resistance. Where do the equivalent resistor's leads attach?