Physics Help Forum Resistance combinations help

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 Apr 30th 2018, 02:34 PM #21 Junior Member   Join Date: Apr 2018 Posts: 13 for the other way i guess that you use the potential divider (R5/R3+R5)*6 For the opamp part i dont have that much of a difficulty yes i know how to calculate each gain inverting summing buffer i just need to check the answers
Apr 30th 2018, 04:27 PM   #22
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 1) You can use this voltage to calculate the voltage across R5 in my fig 4 which is therefore the voltage across R1 and R2 in my fig 2 (There is also another way, can you spot it ?)

 for the other way i guess that you use the potential divider (R5/R3+R5)*6

The current through R3 is also the current through R5 and you know both these so Ohm's Law (again) for the voltage across R5.

But your voltage divider would also work, but be careful they are more tricky when there are many branches / loops about.

 For the opamp part i dont have that much of a difficulty yes i know how to calculate each gain inverting summing buffer i just need to check the answers
Well you clearly got the gain right for A2, but did you realise that the gain is different for each input to A1 and A3?

I hope putting these circuits under the microscope as we have done has helped you learn something. They are both good teaching circuits if you are prepared to look and see what is going on as well as just applying formulae.

 May 3rd 2018, 01:17 AM #23 Junior Member   Join Date: Apr 2018 Posts: 13 thank you buddy the explanation greatly help me for the exams especially for the capacitance question where i worked systematically as you have showed me do you have any idea how to solve this one?
 May 3rd 2018, 02:36 AM #24 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Well that's good and since it was worth 20 marks, I hope you remembered that series and parallel combinations for capacitors works the other way round than for resistors. I will look at the second question but basically you need to find the time rate of change of flux from (the time rate of change of area, plus the time rate increase of flux) and equate that to the emf generated. Dividing this by the resistance will give the current at t = 2. Note you are given the resistance, not the resistivity. Will this change with time? What would happen if you were given the resistivity?
May 4th 2018, 07:35 AM   #25
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 Originally Posted by studiot Well that's good and since it was worth 20 marks, I hope you remembered that series and parallel combinations for capacitors works the other way round than for resistors. I will look at the second question but basically you need to find the time rate of change of flux from (the time rate of change of area, plus the time rate increase of flux) and equate that to the emf generated. Dividing this by the resistance will give the current at t = 2. Note you are given the resistance, not the resistivity. Will this change with time? What would happen if you were given the resistivity?
yes i did remember that for capacitance it was the inverse of resistors
and resistivity does change change for a material at constant T but resistance does vary depending on X-sect
if we were given the resistivity i think we would have to get the X-sect area at the required time then (i would assume they would give the lenghts) use the formula relating resistance is resistivity*L/Area.

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