Question

A company makes two electronic circuit boards that require the same resistance. The Pro-X board runs on 120 volts, while the Pro-I board runs on 90 volts. If the current running through the Pro-I board is 15 fewer milliamps (mA) than the current running through the Pro-X board, how much current is running through each board? Which of the following rational equations bests models this situation?

Answer 1

Answer:

60 mA and 45 mA

$\displaystyle \frac{120}{R}=\frac{90}{R}+0.015$

Step-by-step explanation:

Application of Linear Equations

The voltage V measured in a resistive element of an electrical circuit or resistance R is

$V=I.R$

Where I is the current flowing through the resistor. Solving for I

$\displaystyle I=\frac{V}{R}$

There are two electronic circuit boards that have the same resistance. The Pro-X board runs on 120 volts and the Pro-I board runs on 90 Volts. We know that the current running through this last one is 15 mA fewer than the current through the Pro-X board. It means

$\displaystyle \frac{120}{R}=\frac{90}{R}+0.015$

Rearranging:

$\displaystyle \frac{120}{R}-\frac{90}{R}=0.015$

Operating

$\displaystyle \frac{30}{R}=0.015$

Or, equivalently

$\displaystyle R=\frac{30}{0.015}=2000\Omega$

Thus, the current through the Pro-X board is

$\displaystyle \frac{120}{2000}=60\ mA$

And the current through the Pro-I board is

$\displaystyle \frac{90}{2000}=45\ mA$