Solve {l}{E_{c}=(0.0002)I^2+(0.008)I}{text{when}I=8000}

$$\left. \begin{array} { l } { E _ { c } = ( 0.0002 ) I ^ { 2 } + ( 0.008 ) I } \\ { \text { when } I = 8000 } \end{array} \right.$$

Solve for E_c, I

$$E_{c}=12864$$

$$I=8000$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$E_{c}=0.0002\times 8000^{2}+0.008\times 8000$$

Calculate 8000 to the power of 2 and get 64000000.

$$E_{c}=0.0002\times 64000000+0.008\times 8000$$

Multiply 0.0002 and 64000000 to get 12800.

$$E_{c}=12800+0.008\times 8000$$

Multiply 0.008 and 8000 to get 64.

$$E_{c}=12800+64$$

Add 12800 and 64 to get 12864.

$$E_{c}=12864$$

The system is now solved.

$$E_{c}=12864 I=8000$$