Question

Just need E

A resistor R = 6 Ωis connected in series with an inductor L = 135 mH and a battery V = 12 volts at t = 0.

a) What is the time constant of this LR circuit?

b) How many seconds will it take for the voltage across the inductor to equal 10 volts?

c) How many seconds will it take for the current to increase to 80% of its maximum value?

d) How many seconds will it take for the energy stored in the inductor to increase to 80% of its maximum value? What is its maximum value?

e) After minute, what is the total energy that the battery has supplied, the total energy lost as heat to the resistor

Answers should be:

a) 22.5 msec

b) 4.1 sec

c.) 36.2 msec

d) 50.6 msec, Umax = .27 Joules

e) Ubattery = 1439.46 Joules, Uresistor = 1439.19 Joules, Uinductor = .27 Joules

Answer 1

given :

R = 6 Ω

L = 135 mH = 0.135 H

V = 12

a) time constant = L/R = 0.135 / 6 = 0.0225 s = 22.5 ms [answer]

(b) current as function of time in LR dc circuit is ------(i)

the first derivative of current with respect to time is

voltage across inductor is given by L di/dt = .

12 V is the maximum voltage across inductor.

Let time t is taken by voltage across inductor to reach 10V.

then,

=> t = 0.0041 s = 4.1 ms [answer]

c)

current as function of time in LR dc circuit is

V/R is the maximum current.

let time t is taken by current i to reach 80% of maximum current.

therefore,

=> t = 0.036 s. [answer]

d) Energy stored in inductor is given by .

Maximum value of current we get by putting t = infinity in equation (i) .

.

Maximum energy stored is [answer].

let time taken is t to reach energy stored in inductor 80% of maximum value.

then,

=> t = 0.0506 s = 50.6 ms [answer]