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Thread: 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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    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 $\displaystyle P = 2.5$ and $\displaystyle 10I - 8I^2 = 2.5\Rightarrow -8I^2 + 10I - 2.5 = 0$. This is just a quadratic equation. If we apply the quadratic formula, we get

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

    Alternatively, you can try completing the square:

    $\displaystyle -8I^2 + 10I = 2.5$

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

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

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

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

    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:

    $\displaystyle \left\{\begin{array}{l}
    0.2I_1 + 2(I_1 - I_2) = 5\\
    3I_2 + 0.4I_2 - (I_1 - I_2) = 12\end{array}\right.$

    Simplify:

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

    Multiply (2) by 2.2:

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

    Add (1) and (3):

    $\displaystyle 0I_1 + 7.68I_2 = 31.4$

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

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

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

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

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