Help with algebra equations

**gram1210** · June 16th 2008, 12:12 PM

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

**Reckoner** · June 16th 2008, 12:51 PM

Originally Posted by gram1210

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.

Originally Posted by gram1210

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:

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

Simplify:

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

Multiply (2) by 2.2:

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

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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