# Check my answer for a question regarding Thevenin Theorem

#### Ashenrise

Joined May 2, 2016
2
This is exactly how my professor formatted the question:
For the circuit shown below, calculate:
(a) Thevenin Voltage (Vth)
(b) Thevenin Resistance Rth
(c) Load current ( IL )

this is what I got but I'm not sure what I'm doing wrong:

Vt= 10.9V - 0.9 V= 10 V

1. Vth= R2xVt/(R2+R1)

2. Vth= 600 x 10/600+400= 6000/1000= 6 V

3. Rth= R1+R2= 400+600= 1kohm

4. IL= Vth/RL+Rth= 6/1+1= 3mA

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#### HW-nut

Joined May 12, 2016
97
From your drawing, I am not exactly sure where the load is looking into, so I’m going to assume that the load is R3.

Vth = V1 – V2 = 10V

And Vth = R1+ R2 = 1K

IL = Vth / (Rth + RL) = 10/ (1K + 1K) = 5 mA

Hope this helps.

Last edited:

#### HW-nut

Joined May 12, 2016
97
Whoops, not supposed to answer the question. Let’s try this again.

From your drawing, I am not exactly sure where the load is looking into, so I’m going to assume that the load is R3.

Vth is the open circuit voltage, What is the voltage at R3 if R3 is removed from the circuit?

Rth is the output resistance, what is the resistance across R3 if R3 is removed and the voltage sources are shorted?

Determining the load current should be straightforward once the circuit is redrawn in Theveninequivalent form.

#### Ashenrise

Joined May 2, 2016
2
Whoops, not supposed to answer the question. Let’s try this again.

From your drawing, I am not exactly sure where the load is looking into, so I’m going to assume that the load is R3.

Vth is the open circuit voltage, What is the voltage at R3 if R3 is removed from the circuit?

Rth is the output resistance, what is the resistance across R3 if R3 is removed and the voltage sources are shorted?

Determining the load current should be straightforward once the circuit is redrawn in Theveninequivalent form.
I understand now, thanks so much, I used multisim to check it again and your answers were there, I don't know why I didn't get it the first time. Anyway, I appreciate the help greatly.
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