Thevenin's & Kirchhoff's law, in a circuit

[ahmed markhoos]

I have a circuit, I want to know how much voltage on the 5 ohm's resistor at the far right.

using the two methods of Thevenin's & Kirchhoff's:

Thevenin's: I get 5 V
Kirchhoff's : I get 3.33 V

I don't know what is exactly wrong. but I've NOTICED thing!, when I simulate it in multisim, and connect the voltmeter in series with the far right 5 ohm's, I get 5V. and when I connect it in parallel with the 5 ohm's I get 3.33 V.

I'll try to put my calculations later on.

 

[berkeman]

I have a circuit, I want to know how much voltage on the 5 ohm's resistor at the far right.

using the two methods of Thevenin's & Kirchhoff's:

Thevenin's: I get 5 V
Kirchhoff's : I get 3.33 V

I don't know what is exactly wrong. but I've NOTICED thing!, when I simulate it in multisim, and connect the voltmeter in series with the far right 5 ohm's, I get 5V. and when I connect it in parallel with the 5 ohm's I get 3.33 V.

I'll try to put my calculations later on.

Voltmeters go in parallel and current meters go in series.

 

[ahmed markhoos]

Voltmeters go in parallel and current meters go in series.

I know that, this isn't my question. I'm just surprisingly saying that even that I did it wrong the number is appearing when I do that think. take it as an observation >*<

 

[berkeman]

Can you show your work solving the KCL equation for the top node where R1, R2 and R3 connect?

 

[ehild]

The Thevenin voltage is 5 V but the voltage across the resistor R3 is 3.33 V. The voltmeter connected parallel with R3 measures that voltage. If you connect the voltmeter in series with R3, the voltmeter measures the open-circuit voltage: no current flows through R3, as the resistance of the voltmeter is very high, it can be considered infinity. So the voltmeter reading is equal to the Thevenin voltage.

 

[ahmed markhoos]

Can you show your work solving the KCL equation for the top node where R1, R2 and R3 connect?

by Kirchhoff's: using the upper junction, I1 enter from R1, I2 out to R2, I3 out to R3.

I1=I2+I3

right loop: 5*I3-5-5*I2 =0, SO: I3=1+I2
left loop: 5*I1+5*I2+5-5=0, SO: I1=-I2

by substitution: -I2=I2+1+I2 , THEN: I2=-1/3 AND I3=2/3

BY OHM'S LAW: V=RI= (5)(2/3) = 3.33 V
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by Thevenin's: first we remove R3, give us a circuit with two opposite batteries, so I=0

taking the loop that contain Vth (right one): 0*5 +5 +Vth = 0

Vth=-5.

---------------

The Thevenin voltage is 5 V but the voltage across the resistor R3 is 3.33 V. The voltmeter connected parallel with R3 measures that voltage. If you connect the voltmeter in series with R3, the voltmeter measures the open-circuit voltage: no current flows through R3, as the resistance of the voltmeter is very high, it can be considered infinity. So the voltmeter reading is equal to the Thevenin voltage.

thank you ehild, you are right. the problem is that the doctor didn't explain thevenin's correctly and didn't mention what Vth exactly is. He just gave us the method.thank you very much.

 

[LvW]

The most simple way for solving the task is applying the substitution theorem.
We get the same circuit two times: 5*2.5/(2.5+5) + 5*2.5/(2.5+5)=3.333V.