6. -(4 points SerPSE9 32.P.034.WI. My A 10.9-V battery, a 5.05- resistor, and a

Question

6. -(4 points SerPSE9 32.P.034.WI. My A 10.9-V battery, a 5.05- resistor, and a 10.1-H inductor are connected in series. After the current in the circuit has reached its maximum value, calculate the following. (a) the power being supplied by the battery (b) the power being delivered to the resistor (c) the power being delivered to the inductor (d) the energy stored in the magnetic field of the inductor Need Help? L-ul Loenr 7· · 1 points SerPSED 32.P 045. On a printed circuit board, a relatively long, straight conductor and a conducting rectangular loop lie in the same plane as shown in the figure below. Taking h 0.400 mm, w = 1.00 mm, and l = 2.80 mm, find their mutual inductance. pH Need Help? Resdit

Explanation / Answer

Given,

V = 10.9 V ; R = 5.05 Ohm ; L = 10.1 H

a)The power will be:

P = V^2/R = 10.9^2/5.05 = 23.53 W

Hence, P = 23.53 W

b)P = V^2/R = 10.9^2/5.05 = 23.53 W

Hence, P = 23.53 W

c)The drop at the inductor will be zero So.

P = 0.

d)Stored energy will be:

U = 1/2 L I^2

U = 0.5 x 10.4 x (10.9/5.05)^2 = 23.53 J

Hence, U = 23.53 J

[kind request to put one question at a time. Kindly check and let me know for any query in comment. Rate it]

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