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Inductor in simple circuit w/ graph Problem Help

  • Thread starter SWFanatic
  • Start date
12
0
1. Homework Statement
In Figure 30-63, the inductor has 26 turns and the ideal battery has an emf of 16 V. Figure 30-64 gives the magnetic flux through each turn versus the current i through the inductor. If switch S is closed at time t = 0, at what rate di/dt will the current be changing at t = 1.9L?
Figure 30-63 is a simple circuit with a battery, resistor, switch, and inductor.
Figure 30-64 is a flux vs. current graph, with a constant slope of 2E-4H.


2. Homework Equations
L= flux/ current
emf = L|di/dt|
emf = |d(flux)/dt|


3. The Attempt at a Solution
I am rather lost. The graph shows flux per turn over current, but how do I know what the flux of each turn is? The graph is just an increasing slope with seemingly no boundries. Also, isnt the slope just (d(flux)/dt)/(di/dt)? And if so, wouldnt (di/dt) just be 1, or have I forgotten how to read graphs? Thanks for any help.
 

Answers and Replies

454
0
I can't see any pictures
 

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