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A constant voltage of 6.00 V has been observed over a certain time interval across a 3.00 H inductor. The current through the inductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval?

Respuesta :

Answer:

The time interval is 2.5 sec

Explanation:

Given:

applied voltage [tex]V = 6[/tex] V

Inductance [tex]L = 3[/tex] H

Current change [tex]dI = 8 -3= 5[/tex] A

From the formula of magnitude of induced emf in terms of inductance,

   [tex]V = L\frac{dI}{dt}[/tex]

For finding the time interval,

   [tex]dt = \frac{L dI}{V}[/tex]

   [tex]dt = \frac{3 \times 5}{6}[/tex]

   [tex]dt = 2.5[/tex] sec

Therefore, the time interval is 2.5 sec

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