Question

(17%) Problem 3: An inductor with L = 4.5 H is in a closed circuit with I = 0.85 A (see the diagram). When the switch is opened, the current goes to zero in ?t = 0.19 s. Randomized Variables L = 4.5 H I = 0.85 A ?t = 0.19 s 33% Part (a) Express the magnitude of the average induced emf on the inductor, ?ave, in terms of L, I and ?t. ?ave = 33% Part (b) Calculate the numerical value of ?ave in volts.

          (17%) Problem 3: An inductor with L = 4.5 H is in a closed circuit with I = 0.85 A (see the diagram). When the switch is opened, the current goes to zero in ?t = 0.19 s.

Randomized Variables
L = 4.5 H
I = 0.85 A
?t = 0.19 s

33% Part (a) Express the magnitude of the average induced emf on the inductor, ?ave, in terms of L, I and ?t.
?ave =

33% Part (b) Calculate the numerical value of ?ave in volts.

(17%) Problem 3: An inductor with L = 4.5 H is in a closed circuit with I = 0.85 A (see the diagram). When the switch is opened, the current goes to zero in ?t = 0.19 s.

Randomized Variables
L = 4.5 H
I = 0.85 A
?t = 0.19 s

33% Part (a) Express the magnitude of the average induced emf on the inductor, ?ave, in terms of L, I and ?t.
?ave =

33% Part (b) Calculate the numerical value of ?ave in volts.

Added by Carol D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Cyra Jelle Calleja

02/23/2022

Step 1

85 A in this case) \( \Delta t \) = time taken for the current to change (0.19 s in this case) Show more…

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Problem 3: An inductor with L = 4.5 H is in a closed circuit with I = 0.85 A (see the diagram). When the switch is opened, the current goes to zero in Δt = 0.19 s. Randomized Variables: L = 4.5 H I = 0.85 A Δt = 0.19 s Part (a) Express the magnitude of the average induced emf on the inductor, εave, in terms of L, I and Δt. Part (b) Calculate the numerical value of εave in volts.