10.A 10.7-V battery, a 4.95- resistor, and a 9.1-H inductor are connected in ser

Question

10.A 10.7-V battery, a 4.95- resistor, and a 9.1-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 ___W

(b) the power being delivered to the resistor ____W

(c) the power being delivered to the inductor ____W

(d) the energy stored in the magnetic field of the inductor J

11An emf of 111.5 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.00 A/s. What is the mutual inductance of the two coils?
_____mH

Explanation / Answer

a )

given

V = 10.7 volts

R = 4.95 ohms , L = 9.1 H

a )

the power being supplied by the battery P is

P = V2 / 2R

P = 10.7 X 10.7 / 2 X 4.95

P = 23.129 W

b )

the power being delivered to the resistor is same so

P = V2 / 2R

P = 10.7 X 10.7 / 2 X 4.95

P = 23.129 W

c )

the power being delivered to the inductor is P = 0

d )

the energy stored in the magnetic field of the inductor is E

E = 1 / 2 L X i2

but V = i R ( according to ohm's law )

10.7 = i X 4.95

i = 2.16 A

E = 0.5 X 9.1 X 2.16 X 2.16

E = 21.22 J

-------------------------------------------------

11 )

the mutual inductance of the two coils is = e X ( di/dt )

here e = emf = 111.5 mV = 111.5 X 10-3 V

di/dt = 1 A/s

the mutual inductance of the two coils is = 111.5 X 10-3 X 1

= 111.5 X 10-3 H

= 111.5 mH

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