Problem with voltage divider.

[hatela]

Dear Friends,
I have a problem concerning voltage divider. As you all can see in the picture,

all resistors are 10K. Vin=30volts and o/p is across r3.
Now when I do simple KVL I get the volt drop across R1 as 20Volts and across R3 and R2 as 10 Volts each. So is tyhe o/p voltage 10 volts.. ????
And if I find using thevenin theorem, I get the vth as around 10 volts. Rth as 3.33K ohms and the short circuit current as 3mA. So according to the thevenin equivalent circuit comprising of the 10 volt battery and 3.33Kohm resistoe in series with the 10K ohm load the load voltage comes to around 5 Volts. (approx=4.5 volts).
Can anyone please tell me what is my mistake or where did I go wrong.....

Added after 34 minutes:

oops... sorry, the o/pvoltage using thevenins comes to 7.5 volts....

 

[v_c]

\[ V_{out} = \frac{R_2||R_3}{R_1 + R_2||R_3} V_{in}\]
So you get
\[ V_{out} = \frac{5k\Omega}{10k\Omega + 5k\Omega} 30V=10V\]
This is in fact the Thevenin voltage. Thevenin equivalent circuit is 10V in series with a resistance of \[R_{th}=R_1||R_2||R_3=3.33k\Omega\]. The output voltage is the open circuit voltage, so once you find the Thevenin voltage, you do not need to do any additional voltage divider calculations. Since the output is not loaded the current through the Thevenin resistance is zero and there is no voltage across it. That is why the Thevenin voltage and the output voltage are one and the same in this case. If you had a load resistance to the right of the output terminals, than the output voltage would be less than the Thevenin voltage. Does this help?

Best regards,
v_c

 

[sxunxs]

\[ V_{out} = \frac{R_2||R_3}{R_1 + R_2||R_3} V_{in}\]
So you get
\[ V_{out} = \frac{5k\Omega}{10k\Omega + 5k\Omega} 30V=10V\]
This is in fact the Thevenin voltage. Thevenin equivalent circuit is 10V in series with a resistance of \[R_{th}=R_1||R_2||R_3=3.33k\Omega\]. The output voltage is the open circuit voltage, so once you find the Thevenin voltage, you do not need to do any additional voltage divider calculations. Since the output is not loaded the current through the Thevenin resistance is zero and there is no voltage across it. That is why the Thevenin voltage and the output voltage are one and the same in this case. If you had a load resistance to the right of the output terminals, than the output voltage would be less than the Thevenin voltage. Does this help?

Best regards,
v_c

simple math...

 

[v_c]

Of course, it is simple math! The original poster's problem was not with math. It was not realizing that the thevenin equivalent voltage source is the open circuit voltage for his circuit. All the numbers were already calculated but he did an extra (unnecessary) voltage divider calculation which gave the wrong answer.

Best regards,
v_c

 

[hdrzaman]

So you get

This is in fact the Thevenin voltage. Thevenin equivalent circuit is 10V in series with a resistance of . The output voltage is the open circuit voltage, so once you find the Thevenin voltage, you do not need to do any additional voltage divider calculations. Since the output is not loaded the current through the Thevenin resistance is zero and there is no voltage across it. That is why the Thevenin voltage and the output voltage are one and the same in this case. If you had a load resistance to the right of the output terminals, than the output voltage would be less than the Thevenin voltage