Find the thevenin equivalent of these circuits?

Question

I've tried it and got Vth = 35.9V and Rth=17.4k ohm for (a). Am I right? If not can you explain how its done pls? thanks

http://i292.photobucket.com/albums/mm37/NoM_2008/Untitlfed.jpg

Answer 1

For the left one, you calculate voltage out with RL open. Then calculate current out with RL shorted. Divide the two to get thevenin resistance.

It's drawn incorrectly, the RL should be outside the output points.

Vth, it's just a voltage divider across the battery

Vth = 25*75/ (75+100+150)

Vth = 5.77 volts

To calculate current, first take R4 in parallel with R2

R42 = 25*75/100 = 18.8 ohms

that makes a new voltage divider from 25v

V42 = 25*18.8 / (18.8+250) = 1.749 volts

That voltage cause current to flow thru the 25 ohms

I4 = 1.749 / 25 = 0.0699 amps (this is short circuit current)

Rth = 5.77 / 0.0699 = 82.5 ohms

Edit, had a few mistakes, corrected.

you can also model it as a current source of 0.0699 across a 82.5 ohm resistor.

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The right one:

Voltage with RL open:

Equivalent R at R1 is R1 in parallel with R2,3,4 in series, or 15 ohms in parallel with 12 ohms

R = 15*12/27 = 6.67 ohms

Voltage across that is

V = 9*6.67 = 60 volts

60 volts drives a voltage divider of R2,3,4

V4 = 60*4/15 = 16 volts = Vth

Current with RL short

Equivalent R at R1 is R1 in parallel with 11 ohms

R = 11*12/23 = 5.74 ohms

Voltage is 9*5.4 = 51.7 volts.

Current out is 51.7/11 = 4.70 amps.

Rth = 60/4.7 = 12.8 ohms

So this can be modeled by 60 volts in series with 12.8 ohms, or with 4.7 amps through 12.8 ohms.

If I didn't make another mistake, it's been a while since i did one of these.

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