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Please have a look onExample 4.8. There are two scans.

1: http://img585.imageshack.us/img585/6673/understandingthevenin1.jpg

2: http://img818.imageshack.us/img818/3301/understandingthevenin2.jpg

First we find the equivalent resistance of the circuit on the left of terminals "a" and "b". The equivalent resistance comes to be 4Ω. In other words, we can replace the circuit part lying on left of terminals "a" and "b" with only one resistor of 4Ω. I hope I have it right.

Then we find Vth voltage which is appearing around terminals "a" and "b". This is the voltage which can be detected by a voltmeter around the terminals a-b. Vth is 30V.

Now have a look on Figure 4.29 in scan #2. We have 4Ω resistor (Rth, or equivalent resistance) in series with 30V DC source and RL (load resistor) which is connected to terminals a-b. It is obvious that some of the Vth, 30V, is going to drop around 4Ω resistor which means now the voltage which will appear around the terminals "a" and "b" will be less than 30V. Originally we found the Vth as the voltage which appears around the terminals a-b but now voltage which is appearing on terminals a-b is not Vth. Why is so? I hope you understand my question. Could you please tell me? Thanks.

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# Vth voltage in thevenin theorem

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