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Disclaimer i am sure this is super simple, but i am a complete novice to electronics.

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I am trying to create thevenins equivalent circuit. For VTH i have been calculating IOC which I believe to be, 12(27/77) = 4.2v

However I am debating whether I should be including the 1 ohm resistor in my calculation.

Next for RTH I have shorted the source and opened the load, to calculate RTH I have thus done 1 + (50 * 27/50 + 27) = 18.53ohms

To clarify, VTH = 4.2v and RTH = 18.53ohms

Nortons. What I have been doing to calculate I nortons is to divide the source voltage by any of the resistors on the route to the short circuit in this case just the 50 ohm.

IN = ISC = 12/50 = 0.24A

Now here is my problem, I have read that VTH = IN * RN so 4.2 = 0.24 * 18.53 which of course isn't correct, 0.24 * 18.53 = 4.4472

I am failing to see where I have gone wrong!

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  • \$\begingroup\$ check my answer and do click on the tick icon to mark the answer complete and close this. \$\endgroup\$ – Ashik Anuvar Nov 19 '15 at 10:10
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For Rth while calculating you have gone wrong.

1 ohm and 50 ohms come in series which together come parallel to 27 ohms.

So, RTh = 51 || 27 = 17.65 ohms

Vth = 12/78 * 27 = 4.15 volts.

Now in Nortons theorem,

When you short the load the 27 ohm also is effectively removed.

Isc = 12/51 =0.235 amps.

Rth = 17.65 ohms.

Isc x Rth = 4.15 which is our VTh.

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  • \$\begingroup\$ ah thank you so much! i've been stuck for a few days but you've made it really clear where i went wrong. \$\endgroup\$ – edin1232 Nov 19 '15 at 10:21

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