Current flowing through the resistor

[UltrafastPED]

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.

 

[joaofbi1]

But where exactly did you get 0.55mA from? That has completely stumped me.

 

[UltrafastPED]

Current I2 from message #28.

 

[joaofbi1]

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

 

[joaofbi1]

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?

 

[UltrafastPED]

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!

 

[joaofbi1]

I still don't understand but thank you to everyone that helped me.

 

[gneill]

Would you like to go through it again with a slightly different approach?

 

[joaofbi1]

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