Question

3) A solenoid with 31 x 103 turns per meter, a cross sectional radius of 0.4 m, and a length of 0.2 meters is used as to form an inductor. Then a 9 x 103 volt battery (DC) is connected in series to this inductor, a resistor of 3 kilo-Ohms, and another resistor of 7 kilo-Ohms. After 13 milli-seconds, what is the power output of the battery in kilo-watts?

Answer 1

First of all, we find the the inductance of this solenoid

L = uo N2 A / L

where N = nL ( n is turns / meter)

so,

L = uo n2 L A

L = 4e-7 * 310002 * 0.2 * pi * 0.42

L = 121.4 H

Now,

for series LR circuit, time constant

T = L/R

and

I = Io ( 1 - e- t / T )

R = 7000 + 3000 = 10,000 ohms ( series)

so,

T = L / R = 121.4 / 10000 = 0.012414 sec

so,

P = VI

P = V * Io ( 1 - e- t / T )

P = V * V / R ( 1 - e- t / T )

P = V2 / R ( 1 - e- t / T )

P = 90002 / 10000 ( 1 - e-13e-3 / 0.01214 )

P = 90002 / 10000 * ( 0.657)

P = 5323.96 Watts

OR

P = 5.323 Kilo - watts