Finding the current in resistors

AI Thread Summary
The discussion focuses on calculating the current through resistor R1 in a circuit with given resistances and a 6.00 V battery. The Loop Rule is applied to establish the relationship between the voltages and resistances, leading to the equation E - i(R1 + Req) = 0. The equivalent resistance (Req) of R2, R3, and R4 is calculated as 18.609 ohms, resulting in a total resistance of 117.6 ohms when combined with R1. The initial calculation of current resulted in an incorrect value due to a misplaced decimal point, with the correct current being approximately 0.051 A. The discussion highlights the importance of careful calculations in circuit analysis.
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Homework Statement


In the figure, R1 = 99 . R2 = R3 = 49.5 . R4 = 75.0 , and the ideal battery has emf script e = 6.00 V.

Re-GLORY.gif


What is i in R1?

Homework Equations


V=IR
Junction Rule
Loop Rule?
i = E / (R+r)


The Attempt at a Solution


The Loop Rule states that for any loop in a circuit, the sum of the voltages across the things in it will be zero. So by combining R2 R3 R4 into Req of resistance 18.609 ohms, I have a loop, so I add the voltages:

E - iR1 - iReq = 0
E = 6 as given in the problem. (Right? Maybe I'm wrong here.)
By the Junction Rule, the current i at R1 and Req would be the same.
So,
6 - i(R1 + Req) = 0
But this just gives me the equivilant resistance for all the resistors, and let's call that plain R.
So I have i = E/R, which is an equation I already had, so I felt good about coming across it via application of the two rules, but when I use this to find the current I don't get the right answer. My numerical answer to the question is 0.51016494 A.

I don't exactly know where I'm going wrong here...

Thank you for looking at this.
 
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I get the 18.6 ohms, too. Adding R1 in series, we have 117.6 ohms.
I = V/R = 6/117.6 = .051 A. Looks like you misplaced the decimal point. Quite forgivable at this time of night.
 
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