- #1
clairez93
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Homework Statement
See image attached. Please ignore all work you see around the problem. :]
I need help with 9, 10, and 11 in this picture.
Homework Equations
V = IR
The Attempt at a Solution
9. [tex]R_{eq} = R + (\frac{1}{2R} + \frac{1}{2R})^{-1} = 2R = 24[/tex]
[tex]V' = (0.5)(\frac{1}{2R} + \frac{1}{2R})^{-1} = 6 V[/tex] (this is the voltage drop across the parallel set of resistors of 2R and 2R)
[tex]I_{2} = I_{3} = \frac{6}{2R} = 0.25[/tex] (this is the current across each of the parallel branches, and they are only equal because they have the same resistance and voltage]
[tex]I_{1} = I_{2} + I_{3} = .25 + .25 = .5[/tex]
[tex]V_{i} = I_{1}R = (.5)(12) = 6[/tex]
[tex]E = V' + V_{i} = 6+6 = 12[/tex]
Correct answer is actually 24 V.
10. [tex]R_{eq} = (\frac{1}{250} + \frac{1}{300})^{-1} = 136.364[/tex]
[tex]V = IR_{eq}[/tex]
[tex]24 = I(136.364)[/tex]
[tex]I = 0.176[/tex]
[tex]V' = (0.176)(136.364) = 24[/tex] (this part is weird, is it because the whole circuit's in parallel anyway so there's really no voltage drop across anything?)
[tex]I = \frac{V}{R} = \frac{24}{300} = 0.08 A[/tex]
Correct answer is actually 40 mA.
11. Taking both the loops clockwise:
[tex]0 = -10I_{1} -20I_{2} - 5I_{1} + 50[/tex]
[tex]0 = 20I_{2} - 10I_{3} - 40[/tex]
Stuck from here.
Any help or pointers would be appreciated.