The current I in an electrical conductor varies inversely as the resistance R of the conductor. The current is 50 amps when the resistances is 20 ohms. What is the current when the resistance is 60 ohms?

I = k/R

50 = k/20, so k = 1000, making

I = 1000/R
I(60) = 1000/60 = 16.67

if y varies directly as x and y = 9 when x =2 find when x =10

E = I*R = 50 * 20 = 1000 Volts = Supply

voltage.

I = E/R = 1000/60 = 16.67 Amps.

To solve this problem, we can use the concept of inverse variation. Inverse variation states that when two variables are inversely proportional, their product remains constant.

Let's designate the current as I and the resistance as R. We can define the relationship between I and R as follows:

I * R = k

Where k is a constant.

We are given that the current is 50 amps when the resistance is 20 ohms. Using this information, we can plug these values into the equation:

50 * 20 = k
1000 = k

Now we can use this value of k to find the current when the resistance is 60 ohms. Let's denote the current as I2:

I2 * 60 = 1000

To isolate I2, we can divide both sides of the equation by 60:

I2 = 1000 / 60
I2 = 16.67 amps

Therefore, when the resistance is 60 ohms, the current in the conductor is approximately 16.67 amps.