The current I in an electrical conductor varies inversely as the resistance R of the conductor. The current is 9 amperes when the resistance is 940 ohms. What is the current when the resistance is 580 ohms?
Thank you yet again. Almost through with this part of the class thank goodness!
i = k/r
9 = k/940
k = 8460
i = 8460/r
i = 8460/580
i = 14.6 amps
by the way, k is the voltage V
E = I*R = 9 * 940 = 8460 V. = Supply voltage.
I =E/R = 8460/580 = 14.6 Amps.
To solve this problem, we can use the concept of inverse variation and set up a proportion.
Inverse variation states that when two variables, such as current (I) and resistance (R), are inversely proportional, their product remains constant.
From the given information, we know that when the current is 9 amperes, the resistance is 940 ohms. Let's denote this relationship as I₁ and R₁.
We can write the equation for inverse variation as:
I₁ × R₁ = I₂ × R₂
where I₂ is the current when the resistance is 580 ohms, and R₂ is the unknown resistance.
Plugging in the given values, we get:
9 × 940 = I₂ × 580
Now, solve for I₂:
I₂ = (9 × 940) / 580
Using a calculator, we find that I₂ is approximately 14.61 amperes.
Therefore, when the resistance is 580 ohms, the current is approximately 14.61 amperes.