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Math Help - Help with algebra equations

  1. #1
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    Help with algebra equations

    Hi, I'm having trouble doing two equations. I have no idea where to start on (or how to go about doing) the 1st one, and the 2nd - well I've endlessly tried many different ways, however it always ends up coming out wrong!!

    The first is:

    The power developed in an electrical circuit is given by P=10I - 8I^2, where I is the current in Amps. Determine the current necessary to produce a power of 2.5 watts in the circuit.

    The second is:

    Applying Kirchoff's laws to an electrical circuit produces the following equations:
    5=0.2I1+2(I1-I2)
    12=3I2+0.4I2-(I1-I2)

    Any help would be really appreciated

    Thanks
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  2. #2
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    Quote Originally Posted by gram1210 View Post
    The power developed in an electrical circuit is given by P=10I - 8I^2, where I is the current in Amps. Determine the current necessary to produce a power of 2.5 watts in the circuit.
    So we have P = 2.5 and 10I - 8I^2 = 2.5\Rightarrow -8I^2 + 10I - 2.5 = 0. This is just a quadratic equation. If we apply the quadratic formula, we get

    I = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{-10\pm\sqrt{100 - 4(-8)(-2.5)}}{2(-8)}

    Alternatively, you can try completing the square:

    -8I^2 + 10I = 2.5

    \Rightarrow I^2 - \frac54I = \frac{2.5}{-8} = -0.3125

    \Rightarrow I^2 - \frac54I{\color{red}\;+ \left(\frac{5 / 4}2\right)^2} = -0.3125 {\color{red}\;+ \left(\frac{5 / 4}2\right)^2}

    \Rightarrow I^2 - \frac54I + \frac{25}{64} = -0.3125 + \frac{25}{64}

    \Rightarrow \left(I - \frac58\right)^2 = \frac{25}{64} - 0.3125

    If you forgot how to complete the square or solve quadratic equations, you will really need to do some review.

    Quote Originally Posted by gram1210 View Post
    The second is:

    Applying Kirchoff's laws to an electrical circuit produces the following equations:
    5=0.2I1+2(I1-I2)
    12=3I2+0.4I2-(I1-I2)
    We have two equations with two unknowns. Just solve the system:

    \left\{\begin{array}{l}<br />
0.2I_1 + 2(I_1 - I_2) = 5\\<br />
3I_2 + 0.4I_2 - (I_1 - I_2) = 12\end{array}\right.

    Simplify:

    \left\{\begin{array}{rcrcrl}<br />
2.2I_1 & - & 2I_2 & = & 5 & \quad{\color{red}(1)}\\<br />
-I_1 & + & 4.4I_2 & = & 12 & \quad{\color{red}(2)}<br />
\end{array}\right.

    Multiply (2) by 2.2:

    -2.2I_1 + 9.68I_2 = 26.4\quad{\color{red}(3)}

    Add (1) and (3):

    0I_1 + 7.68I_2 = 31.4

    \Rightarrow I_2 = \frac{31.4}{7.68} = \frac{785}{192}\approx4.0885

    Substituting \frac{785}{192} for I_2 in (1), we get

    2.2I_1 - 2\left(\frac{785}{192}\right) = 5

    \Rightarrow I_1 = \frac{575}{96}\approx5.9896

    Again, if you don't know how to solve systems of linear equations, you will need to review.
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