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.