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Current flowing through the resistor

  • Thread starter joaofbi1
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  • #26
UltrafastPED
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It should be 3.3 mA through 10k ohms. Pay attention to your units!

And that is the total current; we previously calculated the current for this branch - what is it?
 
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  • #27
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So how do I calculate how much mA I've lost through the resistor? What formula?
 
  • #28
UltrafastPED
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The second to last circuit had two parallel resistors: 2.2k and 10.8 k. Each sees 6 V, so you can apply Ohm's law to each resistor.

I1 = 6 V/2.2k ohms = 2.73 mA; I2 = 6 V/10.8 k ohms = 0.55 mA. These add up to 3.3 mA total current which we obtained from the final circuit.

So the current through the 10 k ohm resistor will be I2 = 0.55 mA.

This is the current that will be available to the parallel resistors 1 k and 3.3 k ohm; when they were combined they provided the addition 0.8 k ohms that was combined with the 10 k resistor in series to obtain the 10.8 k ohm equivalent resistor.

This process of working backwards, and marking the resulting currents on each as we walked backwards up the circuits has provided all of the information needed.
 
  • #29
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It should be 3.3 mA through 10k ohms. Pay attention to your units!

And that is the total current; we previously calculated the current for this branch - what is it?
So how do I Calculate how much mA I have lost through the resistor?
 
  • #30
UltrafastPED
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Note that the voltage seen by these parallel resistors is the same, and can be found as V= 6 V - 10k Ohm x 0.55 mA = 6 - 5.5 [V] = 0.5 V.

Or we could have solved from the equivalent resistance of the parallel resistors which was 770 Ohms, with a current of 0.55 mA gives V = 0.770 kOhms x 0.55 mA = 0.44 [V]

These voltages are close (0.5, 0.44) but we should get the same answer both ways ... this indicates that there is an error earlier in the calculations which you should recheck.

Assuming that 0.5 V is correct you can use Ohm's law to get the current I3 = 0.5 V / 3.3 k Ohms = 0.15 mA.
 
  • #31
UltrafastPED
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So how do I Calculate how much mA I have lost through the resistor?
No current is lost through a resistor; only the voltage is used up.
 
  • #32
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But where exactly did you get 0.55mA from? That has completely stumped me.
 
  • #33
UltrafastPED
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Current I2 from message #28.
 
  • #34
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Could I not just simply do 6V/10kΩ to get 0.6 instead of doing
I2 = 6 V/10.8 k ohms = 0.55 mA
 
  • #35
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So once we know that the current going through the 10k Ohms resistor is 0.55mA how do we then do it?

Shall I convert the 0.55mA into volts then do the same operation as above?
 
  • #36
UltrafastPED
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No ... you need to use the equivalent resistance for this branch, which is 10.8 k.

Go back through and draw each of the reduced circuits, labeling them 0 for the original, 1 for the first reduction, etc until you get to the final reduced circuit which has one resistor, and which provides the total current.

Record the equivalent resistances that have been found for each reduced circuit as you move from 0 to N, and then record the voltage drops and currents as you work your way back from N to 0.

When you do this yourself you should get consistent, correct answers.

I'll be offline for a few days due to international travel; Gute Nacht from Germany!
 
  • #37
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I still don't understand but thank you to everyone that helped me.
 
  • #38
gneill
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Would you like to go through it again with a slightly different approach?
 
  • #39
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No use now, I'll be going to bed and it's due in tomorrow. I'll just ask the teacher how to do it. But I think I may have got it on my own anyway by using Vr=(R1/(R1+R2)x6
 

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