Problem 35 - Beginning and Intermediate Algebra

Question

The power in an electrical system varies jointly as the current and the square of the resistance. If the power is 100 watts when the current is 4 amps and the resistance is 5 ohms, what is the power when the current is 5 amps and the resistance is 6 ohms?

Step-by-Step Solution

Answer

In the given electrical system, the power (P) varies jointly with the current (I) and the square of the resistance (R²). Using the initial values of P=100 watts, I=4 amps, and R=5 ohms, we find the constant of variation (k) to be 1. With the new current of 5 amps and resistance of 6 ohms, the new power is \(P' = k \cdot I \cdot R^2 = 1 \cdot (5) \cdot (6^2) = 5 \cdot 36 = 180\) watts.

Step 1: Defining the Relationship

Given that the power (P) in an electrical system varies jointly as the current (I) and the square of the resistance (R²), we can write the equation as: \[P = k \cdot I \cdot R^2\] where k is the constant of variation to be determined.

Step 2: Determine the Constant of Variation k

We are given the initial power, current, and resistance values as P=100 watts, I=4 amps, and R=5 ohms. We can use these values to find the constant of variation (k) by plugging them into our equation: \[100 = k \cdot 4 \cdot (5^2)\]

Step 3: Solve for k

Now we need to solve the equation for k: \[100 = k \cdot (4 \cdot 25)\] \[100 = k \cdot (100)\] \[k = \frac{100}{100}\] \[k = 1\]

Step 4: Apply k to Find New Power

We are given the conditions for which we need to find the new power: I=5 amps, and R=6 ohms. Now we will use the value of k that we found in Step 3 to calculate the new power (P'): \[P' = 1 \cdot (5) \cdot (6^2)\]

Step 5: Calculate P'

Solve for P' in the equation: \[P' = 1 \cdot (5) \cdot (36)\] \[P' = 5 \cdot 36\] \[P' = 180\]

Step 6: Conclusion

When the current is 5 amps and the resistance is 6 ohms, the power in the electrical system is 180 watts.