Voltage Divider Rule question

Discussion in 'General Electronics Chat' started by Xufyan, Sep 13, 2011.

Aug 3, 2010
114
0
Please look at the attached circuit ,

According to Voltage divider rule ,

V1 should be equals to,

10volts x (33k + 68k) / ( 33k + 68k + 100k ) = 5.02volts .

but the circuit says its 2.812 volts

and V2 equals to,

10volts x (68k) / ( 33k + 68k + 100k ) = 3.38volts .

but according to the circuit its 1.641 volts,
please explain where i am wrong

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2. t_n_k AAC Fanatic!

Mar 6, 2009
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Current will flow in the two 47k resistors which will upset the divider relationship.

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3. Tealc Member

Jun 30, 2011
140
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I'm completely new to this but simulating the initial divider circuit in Multisim with just the 100k, 33k and the 68k give the results you have calculated but as soon as I add in the 47k and additional 100k the voltage on my meters changed and when adding the op-amp in they changed again to the figures you found in your simulation.

As a guess I would say that for the bottom meter reading the 47k and 100k are working in parallel with the 68k to change the voltage divider values.

Aug 3, 2010
114
0
can you please drive the equation, how 47k resistors effect the voltage ?

5. t_n_k AAC Fanatic!

Mar 6, 2009
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The attached analysis may help with understanding why the values aren't what you expect.

The difference amplifier effectively loads the voltage divider.

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6. t_n_k AAC Fanatic!

Mar 6, 2009
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The image didn't show that well - I'll try again

• Voltage divider plus diff amp.jpg
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Aug 3, 2010
114
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thanks alot !!!! Great Explanation !!! Awesome !!!!

8. iONic AAC Fanatic!

Nov 16, 2007
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Why wouldn't the 100K in the lower leg play a roll as well, given it connects to ground??

9. t_n_k AAC Fanatic!

Mar 6, 2009
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It does - included in my analysis.

10. iONic AAC Fanatic!

Nov 16, 2007
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Originally you didn't mention it, so I skipped the analysis, I'll read it now!

Thanks