Parallel/series, calculate current drawn

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The discussion focuses on calculating the current drawn in a circuit with multiple resistors. The initial calculations incorrectly treated all resistors as series when some were actually in parallel. The correct approach involves simplifying the circuit by recognizing that two 30K ohm resistors are in parallel, as well as the 40K and 66.7K ohm resistors. After recalculating, the total resistance is found to be approximately 69,375.5 ohms, leading to a corrected current of about 0.000144A. The importance of correctly identifying resistor configurations is emphasized for accurate calculations.
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Homework Statement


http://img267.imageshack.us/img267/8159/untitled3ct1.gif

Find the current drawn.

Homework Equations



V=IR
R1+R2+Rn
1/ (1/R1)+(1/R2)+(1/R3)+(1/R4)

The Attempt at a Solution



R1=20,000ohms
R2=70,000ohms
R3=96,7000ohms
R4=40,000ohms

1/ (1/R1)+(1/R2)+(1/R3)+(1/R4) = 10037.44

10=I(10037.44)
I=.000996A

Where did I go wrong?
 
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Think of the two 30Kohm resistors as being in parallel. Also, think of the 40 and 66.7 kohm resistors as being in parallel. Now, you have simplified that whole middle section into two resistors in series. you can go from here.
 
Simple error, the 30K resistors are in parallel, as are the 40K and 66.7. These in turn are in series and that resultant in parallel with the other two branches.
 
r1=20,000
r2=15,000
r3=25,004.7
r4=40,000

r2&r3 in parallel= 9,375.5

20,000+9,375.7+40,000=69,375.5

10=I(69,375.7)

I=.000144A ??
 
Almost R2 and R3 are in series.
 
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