Prepare for the 5th Class Power Engineering Exam with comprehensive flashcards and multiple-choice questions. Each question includes hints and explanations. Get exam-ready!

Multiple Choice

What would the correct rearrangement of I = E / R yield if solving for R?

To solve for resistance \( R \) in the formula \( I = \frac{E}{R} \), we start by manipulating the equation to isolate \( R \). Initially, we have: \[ I = \frac{E}{R} \] To rearrange for \( R \), we can multiply both sides by \( R \) first, which gives us: \[ I \cdot R = E \] Next, to isolate \( R \), we divide both sides of the equation by \( I \): \[ R = \frac{E}{I} \] This manipulation correctly shows how resistance \( R \) relates to voltage \( E \) and current \( I \). Therefore, the proper formula for \( R \) is indeed \( R = \frac{E}{I} \), confirming that the rearrangement leading to this answer is correct. The other options do not follow the mathematical principles of rearranging this equation. For instance, options suggesting \( R = I / E \) and \( R = E + I \) do not maintain the correct relationship between voltage, current, and resistance as established by Ohm's Law.

To solve for resistance ( R ) in the formula ( I = \frac{E}{R} ), we start by manipulating the equation to isolate ( R ).

Initially, we have:

[ I = \frac{E}{R} ]

To rearrange for ( R ), we can multiply both sides by ( R ) first, which gives us:

[ I \cdot R = E ]

Next, to isolate ( R ), we divide both sides of the equation by ( I ):

[ R = \frac{E}{I} ]

This manipulation correctly shows how resistance ( R ) relates to voltage ( E ) and current ( I ). Therefore, the proper formula for ( R ) is indeed ( R = \frac{E}{I} ), confirming that the rearrangement leading to this answer is correct.

The other options do not follow the mathematical principles of rearranging this equation. For instance, options suggesting ( R = I / E ) and ( R = E + I ) do not maintain the correct relationship between voltage, current, and resistance as established by Ohm's Law.