Recent content by SparkimusPrime

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    Electric field inside a resistor

    A resistor (20 ohm) is made of a very thin piece of metal wire, length = 3mm and diameter = .1mm. Given that it has a potential of 8 volts and .4 amps of current running through it, what is the electric field inside the wire? I know there's no electrostatic equilibrium here, so I can't just...
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    Circuits with Kirchhoff's rules

    So (using my limited knowledge of matrixes) my equations should be solved so they look like this: a = d+e c= a+b c = b+d+e c-b-d-e = 0 e = ((200*d + 40)/80) e - 5/2*d = 1/2 d = ((20*b+300)/200) d - b/10 = 3/2 b = ((200*d-300)/20) b - 10d = 15 c=((-200*d-80)/70) c...
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    Circuits with Kirchhoff's rules

    I'm going to assume I didn't make any mistakes in modeling the circuit. I've done two other problems similar to this and I'm confident that I've ironed out my misconceptions about modeling the circuits, if not my difficulty with the algebra involved in solving said systems of equations.
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    Circuits with Kirchhoff's rules

    Jacob Chestnut: The constants? I have said equations (see above) should my input be of the form: c=((-200*d-80)/70) -200, 80, 70 Or: c + 20/7d = -8/7 1, 20/7, -8/7 (I had some additional help that counseled me to solve for the variables then input the constants into my...
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    Circuits with Kirchhoff's rules

    here are the values I substituted as well as the equations I used in the solve:
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    Circuits with Kirchhoff's rules

    Ok, I took your advice cookie, I tried to enter the equations in the solver (as I'm rusty on my matrixes), here is what I entered: (for convenience I assigned my difficult to enter variables to single letters easily accessible on the calculator keypad: I_1 = a I_2 = b I_3 = c I_5 = d...
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    Circuits with Kirchhoff's rules

    (see attached for circuit diagram) E1 = 40V E2 = 300V E3 = 80V r1 = 200 ohm r2 = 80 ohm r3 = 20 ohm r4 = 70 ohm I've come up with some very ugly equations to express the current flow across various parts of the circuit, as follows: I_1 = I_5 + I_6 I_3 = I_1 + I_2 I_3 =...
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    Calculating Power Loss in a High Voltage Transmission Line

    Yes sorry, read the problem wrong. Seems to be a common failing with this book, using obscure english to mask the real problem. Thanks. Peter
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    Calculating Power Loss in a High Voltage Transmission Line

    A question surprisingly similar to this: https://www.physicsforums.com/showthread.php?t=17219 I = 1000amp V = 700,000volt distance = 100miles Resistence of the wire = .5 ohm / mile The resistence of my line is 50 ohms, original power is 7e5 * 1e3 = 7e8 watts. Final power, due to P =...
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    Re:Statistics: Applications of sampling thoery

    Edit: Moved from statistics - https://www.physicsforums.com/showthread.php?s=&postid=164120#post164120 I'm doing some homework over the break (!) so I don't have access to my usual lines of help. I've hit a wall: I don't really know how to solve a problem like this, from the...
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    Modeling conservation of charge

    I know there is charge present on each plate of a capacitor, as they store charge (by definition). The charges are equal and opposite and it is the electric forces generated by that charge that allows said charge to be stored. Taken as a system and preserving the signs, yes there is no net...
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    Understanding the Charge Distribution in a Coaxial Cylinder System

    Quanta? I'll assume it means amount. So my intuition was correct about the maximum available charge is spread between the cylinders. Calculating the electric field at the outer cylinder: So you are correct, the electric field does extend beyond the outer cylinder and some distance...
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    Understanding the Charge Distribution in a Coaxial Cylinder System

    First I set about to calculate the electric field: Solve for Q with the capacitance of two coaxial cylinders: substituting: Where 'r' is a distance measured from the center. Trying several distances (r = 5cm -> E = 69.2 Volts / meter, r = 2cm [on the internal cylinder] -> 432.8 Volts...
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    Modeling conservation of charge

    Yes, it seems to me that this should be the answer. Maybe it is a misprint, all the evidence seems to point that way. Calculating the charges on the two capacitors: Q_1 = C1 * V_1 = 5600*2 = 11200 Q_2 = C2 * V_1 = 1400*8 = 11200 It seems that the charges on the capacitors are equal...
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    Modeling conservation of charge

    Whoops, double post.I must've pushed the wrong button. Sorry about that, ignore this, see the other thread.
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