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!

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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