Question

A modern house is wired for \(115 \mathrm{~V}\), and the current is limited by circuit breakers to a maximum of 200 . A. (For the purpose of this problem, treat these as \(\mathrm{DC}\) quantities. a) Calculate the minimum total resistance the circuitry in the house can have at any time. b) Calculate the maximum electrical power the house can consume.

Short answer

Answer: The minimum total resistance for the house circuitry is 0.575 Ω, and the maximum electrical power consumption is 23,000 W or 23 kW.

Step-by-step solution

Part a: Find the minimum total resistance

We can use Ohm's Law to find the minimum total resistance in the circuit. Ohm's Law states that Voltage = Current × Resistance or \(V = I × R\). We are given the voltage, \(V = 115 V\), and the maximum current, \(I = 200 A\). We need to find the resistance, \(R\). Rearrange the formula to get the resistance: \(R = \frac{V}{I}\). Now plug in the values: \(R = \frac{115 V}{200 A}\). Calculate the result: \(R = 0.575 \Omega\). So, the minimum total resistance in the house circuitry is \(0.575 \Omega\).

Part b: Calculate the maximum electrical power consumption

To calculate the maximum electrical power consumption, we can use Joule's Law. It states that the power, \(P\) is equal to the voltage, \(V\) multiplied by the current, \(I\): \(P = V \times I\). We have the voltage \(V = 115 V\) and the maximum current \(I = 200 A\). Substitute the values into the formula: \(P = 115 V \times 200 A\). Calculate the result: \(P = 23,000 W\). Thus, the maximum electrical power the house can consume is \(23,000 W\) or \(23 kW\).