Answer:
60 mA and 45 mA
[tex]\displaystyle \frac{120}{R}=\frac{90}{R}+0.015[/tex]
Step-by-step explanation:
Application of Linear Equations
The voltage V measured in a resistive element of an electrical circuit or resistance R is
[tex]V=I.R[/tex]
Where I is the current flowing through the resistor. Solving for I
[tex]\displaystyle I=\frac{V}{R}[/tex]
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
[tex]\displaystyle \frac{120}{R}=\frac{90}{R}+0.015[/tex]
Rearranging:
[tex]\displaystyle \frac{120}{R}-\frac{90}{R}=0.015[/tex]
Operating
[tex]\displaystyle \frac{30}{R}=0.015[/tex]
Or, equivalently
[tex]\displaystyle R=\frac{30}{0.015}=2000\Omega[/tex]
Thus, the current through the Pro-X board is
[tex]\displaystyle \frac{120}{2000}=60\ mA[/tex]
And the current through the Pro-I board is
[tex]\displaystyle \frac{90}{2000}=45\ mA[/tex]
