Question Not Found | Examoogle

Examoogle • User • info@examoogle.com • EE-2024-8821

Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked

Back to Practice Questions

ElectricalMachine

Prev Next

Calculate the reluctance (A/Wb) when the magnetomotive force is 10 A-turns and the flux is 5Wb.

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalMachine

Option A

Quick Summary:

Given: Magnetomotive force (MMF) = 10 A-turns, Magnetic flux ( $\Phi) = 5$ Wb

📐MAMath SolutionDirect Formula

📋 Given

Magnetomotive force (MMF) = 10 A-turns, Magnetic flux ( $\Phi) = 5$ Wb

🔢 Formula Used

R=MMFΦ\mathcal{R} = \frac{\text{MMF}}{\Phi}R=ΦMMF​

🔢 Step-by-Step Solution

Identify given parameters

From the problem statement, the magnetomotive force is $\text{MMF} = 10\text{ A-turns}$ and the magnetic flux is $\Phi = 5\text{ Wb}$ .

MMF=10 A-t,Φ=5 Wb\text{MMF} = 10\text{ A-t}, \quad \Phi = 5\text{ Wb}MMF=10 A-t,Φ=5 Wb

Recall the formula for reluctance

Reluctance $(\mathcal{R})$ in a magnetic circuit is analogous to resistance in an electric circuit and is given by Hopkinson's law.

R=MMFΦ\mathcal{R} = \frac{\text{MMF}}{\Phi}R=ΦMMF​

Calculate reluctance

Substitute the given values into the reluctance formula: $\mathcal{R} = \frac{10}{5} = 2\text{ A/Wb}$ .

R=10 A-turns5 Wb=2 A/Wb\mathcal{R} = \frac{10\text{ A-turns}}{5\text{ Wb}} = 2\text{ A/Wb}R=5 Wb10 A-turns​=2 A/Wb

✅

A is correct because substituting the given values into Hopkinson's Law yields a reluctance of 2 A/Wb.

Core Concepts Used

Click any tag to open in AI Tutor

Hopkinson's Law Magnetomotive Force (MMF) Reluctance in Magnetic Circuits

💡 EXAM TIP

Reluctance $(\mathcal{R})$ in a magnetic circuit behaves exactly like Resistance $(R)$ in an electrical circuit, where $\text{MMF}$ is analogous to Voltage $(V)$ and Flux $(\Phi)$ is analogous to Current $(I)$ according to Ohm's Law ( $V = IR$ ).

Discussion (0)

Loading discussion...

Prev Next

Correct Answer

Related Questions

Discussion (0)