In a 240V, 20A two-wire branch circuit, the maximum allowable voltage drop is 2.4V. Using Ohm's law, what is the maximum conductor resistance?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $24.99Unlock all

Prepare for the PMMI Industrial Electricity Test with quizzes featuring flashcards and multiple-choice questions. Each answer is explained with hints to ensure a thorough understanding. Gear up to excel in your exam!

Multiple Choice

In a 240V, 20A two-wire branch circuit, the maximum allowable voltage drop is 2.4V. Using Ohm's law, what is the maximum conductor resistance?

Explanation:
The main idea is that the voltage drop along the conductor path in a circuit equals the current times the total resistance of that path (V = I × R). To stay within the allowed drop, you solve for the maximum resistance: R_max = V_drop / I. Here, the allowable drop is 2.4 V and the current is 20 A. So R_max = 2.4 / 20 = 0.12 ohms. In a two-wire branch circuit, this 0.12 ohms represents the total loop resistance of the conductors (the path the current travels there and back).

The main idea is that the voltage drop along the conductor path in a circuit equals the current times the total resistance of that path (V = I × R). To stay within the allowed drop, you solve for the maximum resistance: R_max = V_drop / I.

Here, the allowable drop is 2.4 V and the current is 20 A. So R_max = 2.4 / 20 = 0.12 ohms. In a two-wire branch circuit, this 0.12 ohms represents the total loop resistance of the conductors (the path the current travels there and back).

More Multiple Choice Questions
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy