Simple circuit analysis

Discussion in 'General Electronics Chat' started by dan2004, Jun 17, 2009.

1. dan2004 Thread Starter New Member

Jun 16, 2009
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0
Hello all,

It's been a very long time since I was involved in electronics. I was wondering if someone here would be nice enough to analyze this simple circuit and explain to me again how Vref at R4 is 100mV when R2 is set at 10k. I'm still pretty familiar with Ohm's Law for calculating V, I and R across series and parallel circuits but am so rusty at it that I need someone's help to show me the calculations.

Thanks!
Dan

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2. hobbyist AAC Fanatic!

Aug 10, 2008
885
87
I think when R2 is at 10K then R2 is in parrallel with the series resistance R3 and R4.

So total current would be roughly 1mA. Then part of that branchs to the R3 R4 combo, producing around 95.8 mV. across R4.

R2*(R3+R4)/(R2+R3+R4) = 9099 ohms.

That +R1 = 9099 + 5K = 14099

!5V. / 14099 = 1.06 mA. total current.
-------------------------

(KCL)

.00106 * 10K / (10K + 100K + 1K) = 95.49 uA.

95.49 uA * 1K = 95.49 mV.

I believe this to be correct.

Last edited: Jun 17, 2009
3. dan2004 Thread Starter New Member

Jun 16, 2009
3
0
Thank you so much for your reply. The only part I am hazy about is Kirschoffs Law.

I understand the first part:

R2*(R3+R4)/(R2+R3+R4) = 9099 ohms.

That +R1 = 9099 + 5K = 14099

!5V. / 14099 = 1.06 mA. total current

Could you tell me the KCL formula for figuring the current:

.00106 * 10K / (10K + 100K + 1K) = 95.49 uA.

Perhaps if you could show me how you get the current in each branch of the parallel circuit (R2 // R3 +R4)

Thanks!

Last edited: Jun 17, 2009
4. hobbyist AAC Fanatic!

Aug 10, 2008
885
87

The total current times the OPPOSITE branch resistance, divided by the total resistance of both branchs in parrallel, would give you rhe current through the branch of interest.

1.06 mA. total current times the 10K branch divided by the 9099 ohms,
would equal the current through the 101K ohm branch.

The exact opposite of the voltage divider. where it is the toatal voltage times the resistor of interest divided by the total series resistance.

I don't think this is KCL law, I think it's just
"current divider equation."

Last edited: Jun 17, 2009
5. dan2004 Thread Starter New Member

Jun 16, 2009
3
0
Thanks for your help!