Can someone help me redraw this circuit.

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The discussion revolves around redrawing a circuit for clarity on parallel and series connections. The original poster expresses uncertainty about their circuit understanding but later confirms their redrawn circuit is correct. They then seek assistance in calculating the current through resistor 5, given that the current through resistor 6 is 1.64 mA. The current divider rule is suggested as a method to find the current through resistor 5 using its resistance and the equivalent resistance of the circuit. The conversation highlights the importance of practice in mastering circuit analysis.
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man I am not good at doing these circuits.. i do think i have it right however i want to be reassured before i drive this info into my head..

http://galileo.physics.fsu.edu/enc/60/b0693c1d166eb44f829042abcf6390a2cf5a73998caaf05a1d16890408211ac74fb5a1875d078912fd7a599483ded40f7806f73f3857b7e65b1628c75c29fe63f222626d31cef1695b9709989c1a225a

thanks guys.. basically redrawing it so you can see what is parallel and in series..
 
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its asking for a password and username
 
formulajoe said:
its asking for a password and username

oops sorry about that.. here it is
answered my own question lol..

http://img210.exs.cx/img210/2812/resistor6xg.gif
 
yep, that looks right. circuits are a snap with a little bit of practice.
 
OKay i got the right answer but now it is asking for the current through resistor 5. the current through resistor 6 is 1.64 mA.

then for b it wants the potential difference between point a and b.
 
You've got current through resistor 6. Then use current divider rule to get the current through resistor 5 (using R5 and Req3).
 

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