L.R. Circuit Class 12 Notes | EduRev

Physics Class 12

Created by: Infinity Academy

Class 12 : L.R. Circuit Class 12 Notes | EduRev

The document L.R. Circuit Class 12 Notes | EduRev is a part of the Class 12 Course Physics Class 12.
All you need of Class 12 at this link: Class 12

8. L.R. Circuit:

As the switch S is closed in given figure, current in

L.R. Circuit Class 12 Notes | EduRevcircuit wants to rise upto L.R. Circuit Class 12 Notes | EduRev in no time but inductor

opposes it L.R. Circuit Class 12 Notes | EduRev          L.R. Circuit Class 12 Notes | EduRev

hence at time t = 0 inductor will behave as an open circuit

at t = 0

L.R. Circuit Class 12 Notes | EduRev

As the time passes, i in the circuit rises and L.R. Circuit Class 12 Notes | EduRev decreases. At any instant t.

L.R. Circuit Class 12 Notes | EduRev

current reaches the value L.R. Circuit Class 12 Notes | EduRev at time t = ∞ or we can say, inductor will behave as a simple wire.

at t = ∞

L.R. Circuit Class 12 Notes | EduRev

 

Ex. 43 Find value of current i, i1 and i2 in given figure at

L.R. Circuit Class 12 Notes | EduRev

(a) time t = 0

(b) time t = ∞

Ans: (a) At time t = 0 inductor behaves as open circuit

i = v/R

i1 = 0

i2= i = v/R            L.R. Circuit Class 12 Notes | EduRev

(b) At time t = ∞. Inductor will behaves as simple wire

i = L.R. Circuit Class 12 Notes | EduRev

i1 = i2L.R. Circuit Class 12 Notes | EduRev L.R. Circuit Class 12 Notes | EduRev

 

9. GROWTH AND DECAY OF CURRENT IN L-R CIRCUIT:

9.1 Growth of Current:

Consider a circuit containing a resistance R, an inductance L, a two way key and a battery of e.m.f E connected in series as shown in figure. When the switch S is connected to a, the current in the circuit grows from zero value. The inductor opposes the growth of the current. This is due to the fact that when the current grows through inductor, a back e.m.f. is developed which opposes the growth of current in the circuit. So the rate of growth of current is reduced. During the growth of current in the circuit, let i be the current in the circuit at any instant t. Using Kirchhoff's voltage law in the circuit, we obtain

L.R. Circuit Class 12 Notes | EduRev

E - L L.R. Circuit Class 12 Notes | EduRev = R i or E - Ri = L L.R. Circuit Class 12 Notes | EduRev or L.R. Circuit Class 12 Notes | EduRev

Multiplying by - R on both the sides, we get

L.R. Circuit Class 12 Notes | EduRev

Integrating the above equation, we have

loge(E - Ri) = - L.R. Circuit Class 12 Notes | EduRev + A ...(1)

where A is integration constant. The value of this constant can be obtained by applying the condition that current i is zero just at start i.e., at t = 0. 

Hence,

loge E = 0 + A

or A = logeE ...(2)

Substituting the value of A from equation (2) in equation (1), we get

loge(E - Ri) = - L.R. Circuit Class 12 Notes | EduRev + loge E

or logeL.R. Circuit Class 12 Notes | EduRevL.R. Circuit Class 12 Notes | EduRev

or L.R. Circuit Class 12 Notes | EduRev = exp L.R. Circuit Class 12 Notes | EduRev

or 1 - L.R. Circuit Class 12 Notes | EduRev = expL.R. Circuit Class 12 Notes | EduRev

or L.R. Circuit Class 12 Notes | EduRev

Therefore,     i = L.R. Circuit Class 12 Notes | EduRev

The maximum current in the circuit i0 = E/R. 

So,    i = i0 L.R. Circuit Class 12 Notes | EduRev ...(3)

Equation (3) gives the current in the circuit at any instant t. It is obvious from equation (3) that i = i0, when

expL.R. Circuit Class 12 Notes | EduRev = 0 i.e., at t = ∞

L.R. Circuit Class 12 Notes | EduRev

Hence the current never attains the value i0 but it approaches it asymptotically. A graph between current and time is shown in figure.

We observe the following points

(i) When t = (L/R) then

i = i0 L.R. Circuit Class 12 Notes | EduRev = i0 {1 - exp(-1)} = i0 L.R. Circuit Class 12 Notes | EduRev = 0.63 i0

Thus after an interval of (L/R) second, the current reaches to a value which is 63% of the maximum current. The value of (L/R) is known as time constant of the circuit and is represented by t. Thus the time constant of a circuit may be defined as the time in which the current rises from zero to 63% of its final value. In terms of t,

L.R. Circuit Class 12 Notes | EduRev

(ii) The rate of growth of current (di/dt) is given by

L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRevL.R. Circuit Class 12 Notes | EduRev ⇒ L.R. Circuit Class 12 Notes | EduRev ...(4)

From equation (3), exp L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRev

Therefore, L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRev ...(5)

This shows that the rate of growth of the current decreases as i tends to i0. For any other value of current, it depends upon the value of R/L. Thus greater is the value of time constant, smaller will be the rate of growth of current.

Note:

L.R. Circuit Class 12 Notes | EduRev Final current in the circuit = L.R. Circuit Class 12 Notes | EduRev, which is independent of L.

L.R. Circuit Class 12 Notes | EduRev After one time constant, current in the circuit=63% of the final current (verify yourself)

L.R. Circuit Class 12 Notes | EduRev More time constant in the circuit implies slower rate of change of current.

L.R. Circuit Class 12 Notes | EduRev If there is any change in the circuit containing inductor then there is no instantaneous effect on the flux of inductor.

L1i1 = L2i2


Ex. 44 At t = 0 switch is closed (shown in figure) after a long time suddenly the inductance of the inductor is made h times lesser L.R. Circuit Class 12 Notes | EduRev than its initial value, find out instant current just after the operation.

L.R. Circuit Class 12 Notes | EduRev

Ans: Using above result (note 4)

L1i1 = L2i2 ⇒ i2 = L.R. Circuit Class 12 Notes | EduRev


Ex. 45 Which of the two curves shown has less time constant.

L.R. Circuit Class 12 Notes | EduRev

Ans: curve 1


9.2 Decay of Current:

Let the circuit be disconnected from battery and switch S is thrown to point b in the figure. The current now begins to fall. In the absence of inductance, the current would have fallen from maximum i0 to zero almost instantaneously. But due to the presence of inductance, which opposes the decay of current, the rate of decay of current is reduced.

suppose during the decay of current, i be the value of current at any instant t. Using Kirchhoff's voltage law in the circuit, we get

- LL.R. Circuit Class 12 Notes | EduRev or L.R. Circuit Class 12 Notes | EduRev   

    L.R. Circuit Class 12 Notes | EduRev

Integrating this expression, we get

loge i = - L.R. Circuit Class 12 Notes | EduRev + B

where B is constant of integration. The value of B can be obtained by applying the condition that when t = 0, i = i0

Therefore, loge i0 = B

Substituting the value of B, we get

logei = - L.R. Circuit Class 12 Notes | EduRev + logei0

or logeL.R. Circuit Class 12 Notes | EduRev = - L.R. Circuit Class 12 Notes | EduRev

or (i/i0) = expL.R. Circuit Class 12 Notes | EduRev ...(6)

or L.R. Circuit Class 12 Notes | EduRev

where t = L/R = inductive time constant of the circuit.

It is obvious from equation that the current in the circuit decays exponentially as shown in figure.

We observe the following points

(i) After t = L/R, the current in the circuit is given by

i = i0 expL.R. Circuit Class 12 Notes | EduRev= i0 exp(-1)   L.R. Circuit Class 12 Notes | EduRev

= (i0 / e) = i0/2.718 = 0.37 i0

So after a time (L/R) second, the current reduces to 37% of the maximum current i0. (L/R) is known as time constant t. This is defined as the time during which the current decays to 37% of the maximum current during decay.

(ii) The rate of decay of current in given by

L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRev

⇒ L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRevi0 exp L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRev ...(7)

or - L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRev

This equation shows that when L is small, the rate of decay of current will be large i.e., the current will decay out more rapidly.


Ex. 46 In the following circuit the switch is closed at t = 0. Initially there is no current in inductor. Find out current the inductor coil as a function of time.

L.R. Circuit Class 12 Notes | EduRev

Ans: At any time t

- e + i1 R - (i - i1) R = 0

- e + 2i1 R - i R = 0

i1 = L.R. Circuit Class 12 Notes | EduRev

Now, - e + iR + iR +L.L.R. Circuit Class 12 Notes | EduRev = 0 L.R. Circuit Class 12 Notes | EduRev

- e + L.R. Circuit Class 12 Notes | EduRev + iR + i. L.R. Circuit Class 12 Notes | EduRev ⇒ - L.R. Circuit Class 12 Notes | EduRev

L.R. Circuit Class 12 Notes | EduRevdt = - L. di ⇒ L.R. Circuit Class 12 Notes | EduRev

L.R. Circuit Class 12 Notes | EduRev ⇒ L.R. Circuit Class 12 Notes | EduRev

- ln L.R. Circuit Class 12 Notes | EduRev = L.R. Circuit Class 12 Notes | EduRev

i = + L.R. Circuit Class 12 Notes | EduRev


Ex. 47 Figure shows a circuit consisting of a ideal cell, an inductor L and a resistor R, connected in series. Let the switch S be closed at t = 0. Suppose at t = 0 current in the inductor is i0 then find out equation of current as a function of time

L.R. Circuit Class 12 Notes | EduRev

Ans: Let an instant t current in the circuit is i which is increasing at the rate di/dt.

Writing KVL along the circuit, we have

ε - LL.R. Circuit Class 12 Notes | EduRev - iR = 0 ⇒ L.R. Circuit Class 12 Notes | EduRev = ε - iR

⇒ L.R. Circuit Class 12 Notes | EduRev             ⇒ ln L.R. Circuit Class 12 Notes | EduRev = - L.R. Circuit Class 12 Notes | EduRev

⇒ ε - iR = (ε - i0R)e-Rt/L ⇒ i = L.R. Circuit Class 12 Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Dynamic Test

Content Category

Related Searches

Semester Notes

,

Extra Questions

,

pdf

,

video lectures

,

Exam

,

Objective type Questions

,

practice quizzes

,

shortcuts and tricks

,

Free

,

L.R. Circuit Class 12 Notes | EduRev

,

Sample Paper

,

Viva Questions

,

MCQs

,

ppt

,

past year papers

,

mock tests for examination

,

L.R. Circuit Class 12 Notes | EduRev

,

Important questions

,

L.R. Circuit Class 12 Notes | EduRev

,

Summary

,

study material

,

Previous Year Questions with Solutions

;