Recent content by arl146

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    Partial fractions

    ohhhh .. i see what i did. ok .. so i was doing A(3x+ 4)^2 + B(3x+4) not realizing the denominator difference. stupid mistake. thanks!
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    Partial fractions

    1. Homework Statement determine the indefinite integral: ∫ (4x+10)/(9x^2+24x+16) dx 2. Homework Equations partial fractions technique 3. The Attempt at a Solution i know it's partial fractions and i thought i did it right but i got the wrong answer. (4x+10)/(9x^2+24x+16) =...
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    B field intensity of 2 parallel wires

    1. Homework Statement Two long parallel wires are a center-to-center distance of 4.60 cm apart and carry equal anti-parallel currents of 4.40 A. Find the magnetic field intensity at the point P which is equidistant from the wires. (R = 4.00 cm). 2. Homework Equations magnetic field...
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    5 resistor circuit

    what happens if i get a negative value for V? i got V_P' = 3.4413 V and V_P = -3.709497 V may be my calculations but ...i ended up with 0.8477 for Req
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    5 resistor circuit

    ok thanks. when the answers become available ill have to check to see if i would have been right
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    5 resistor circuit

    I think i may have thought that at one point. well my homework was already due so i didn't get it. but still, i have those 7 equations. I can substitute the 5 equations (1 for each I) into the 2 current equations and end up getting VP and VP' so what do I then do with it to find the Req? do i...
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    5 resistor circuit

    would the potential difference across R4 be -V_P? if across R3 it is (8-V_P) and across R5 it is (V_P-V_P') then [(8-V_P)-(V_P-V_P')] is the 'initial' and the final is 8. so [(8-V_P)-(V_P-V_P')]-8 is just -V_P? or is it I4=(V_P'-2*V_P)/R4
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    5 resistor circuit

    what youre saying makes no sense. well i got the first part, but the second part made absolutely no sense to me? i don't understand how to get the I4 equation i dont understand the potential difference across R4
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    5 resistor circuit

    3? well i dont get what the equation would be still with that information :cry: I_4=(V_P-V_P'-8)/R4 so is this wrong then too?
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    5 resistor circuit

    i don't understand why I4 is not right. Isn't it similar to I2?
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    5 resistor circuit

    ummm I_4=(V_P'-V_P+8)/R4? since the potential difference at the right corner just past R2 and R4 is V_P-8? no i dont think thats right .. why isn't it what I had? what's wrong with the 8?
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    5 resistor circuit

    ok well I1 is easy... i assume I3 is the same thing? I_3=(8-V_P')/R3 right? so when you want to do R2 and R4 how do you do it .. is it like I_2=(V_P-8)/R2? and then the same thing for R4 .. I_4=(V_P'-8)/R4 ? if thats the case then I_5=(V_P-V_P')/R5 ? and then the current laws: for P...
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    5 resistor circuit

    nothing in my textbook about it .. professor briefly went over it today in class. theres like 4 formulas that he got. I1R1=Va-Vb I1R1=Vb-Vc I2R2=Va-Vd I2R=Vd-Vc where c is the left corner of the 'diamond' , a is the right corner, b is P in our case, d is P' in our case .. R is our R3 and R3 in...
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    5 resistor circuit

    oh yea true .. well then i guess not .. but that is the original text, all i have in the original post is the whole problem
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    5 resistor circuit

    uhhh i guess? idk! im confused already. isn't that just R5 though?
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