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Time Varying Voltages & Currents 1

Time VaryingVoltages & Currents

1 Electric Oscillations

Suppose you charge a capacitor (capacitance C) and then connect its plates to each other throughan inductor coil (inductance L). What will happen? The process has been depicted in Figure 1.

(t = 7T/8)

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

UE

UE

UE

UE

UE

UE

UE

UE UB

UB

UB

UB

UB

UB

UB

UB

C

C

C

C

C

C

C

C

L

L

L

L

L

L

L

L

(t = 0, t = T )

(t = T/8)

(t = T/4)

(t = 3T/8)

(t = T/2)

(t = 5T/8)

(t = 6T/8)

Figure 1: The free oscillations in an LC circuit for one complete cycle. The bar graphs show the distribution ofelectric and magnetic energy at different instants of time. Note that by the time one complete oscillation of thecharge (or current) has taken place, the energy in either C or L has undergone two oscillations.

Suppose that initially, the upper plate of the capacitor is charged positively and the lower plate

Anant Kumar. RG–6, Armstrong Avenue, Near PUMP HOUSE, Bidhan Nagar, Dgp–12. Mob. No.

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negatively (Figure 1(a)). In this case, the entire energy of the circuit is concentrated in thecapacitor. As soon as the capacitor has been connected to the inductor, it starts to discharge,and a current flows through the coil L. Owing to the opposition by the back emf of the inductor,the current rises slowly. The electrical energy of the capacitor is converted to the magnetic energyof the coil. This process terminates when the capacitor is discharged completely, while the currentin the circuit attains the maximum value (Figure 1(c)) At this instant the entire electrical energyhas been converted into the magnetic energy of the inductor. Starting from this moment, thecurrent starts to decrease retaining its direction. However, it does not ceases immediately sinceit is sustained by the self–induced emf in the inductor. The current recharges the capacitor butthis time the polarity of the capacitor is opposite that of the initial situation.At the same time,the appearing electric field tends to reduce the current. Finally, the current ceases, while thecapacitor is fully charged (Figure 1(e)). The entire energy is once more electrical. From thismoment, the capacitor starts to discharge again, the current flows in the opposite direction, andthe process is repeated (Figure 1(f) through (a)) until the state of the circuit is restored to theinitial state. The entire cycle begins again only to be repeated again and so on.

If the conductors constituting the oscillatory circuit have no resistance, strictly periodic os-cillations will be observed in the circuit. In the course of the process, the charge on the capacitorplates, the voltage across the capacitor and the current in the inductor coil vary periodically. Theoscillations are accompanied by mutual conversion of the energy of electric and magnetic fields.

However, if the resistance of the circuit R �= 0, then, in addition to the process described,electromagnetic energy will be transformed into Joule’s heat. The circuit in this case exhibitsdamped oscillations.

1.1 Equation of an Oscillatory Circuit

Let us consider a circuit containing series connected capacitor C, inductor coil L, resistor R anda time varying external voltage source V (t) (Figure 2). Let us first choose the positive direction

V (t)

1C

q 2

R

L

Figure 2: An LCR circuit

for traversing the loop as clockwise (you could take it as anti–clockwise as well). We denote by qthe charge on the capacitor plate the direction from which to the other plate coincides with thechosen direction of our circumvention of the loop. Then the current in the circuit is defined as

i =dq

dt(1)

Consequently, if i > 0, then dq > 0 as well, and vice versa.In accordance with Ohm’s law for section 1RL2 of the circuit, we have

iR = V1 − V2 + Es + V (t), (2)


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where Es is the self induced emf in the inductor coil. In the case under consideration,

Es = −Ldi

dtand V2 − V1 =

q

C.

Hence Equation 2 can be written in the form

Ldi

dt+ iR +

q

C= V (t), (3)

or taking into account (1),

Ld2q

dt2+ R

dq

dt+

1C

q = V (t). (4)

This is the equation of an oscillatory circuit, which is a linear second order nonhomogeneousdifferential equation with constant coefficients. Using this equation for calculating q(t), we caneasily obtain the voltage across the capacitor as Vc = V2 − V1 = q/C and current i by formula( 1).

The above equation can be given a different form:

d2q

dt2+ 2γ

dq

dt+ ω2

0q =1L

V (t), (5)

where the following notation is introduced:

2γ =R

L, ω2

0 =1

LC. (6)

The quantity ω0 is called the natural frequency of the circuit and γ is the damping factor. Themeaning of these quantities will be explained in due course.

If V (t) = 0, i.e. the circuit contains no source, the oscillations are called free oscillations.They will be undamped for R = 0 and damped for R �= 0. We consider all these cases.

1.2 Free Oscillations

1.2.1 Free Undamped Oscillations

If a circuit contains no external voltage and if its resistance R = 0, the oscillations in such acircuit will be free and undamped. The equation describing these oscillations is a particular caseof Equation 5 when V (t) = 0 and R = 0:

d2q

dt2+ ω2

0 q = 0. (7)

The solution of this equation is the function

q = q0 cos(ω0t + α), (8)

where q0 is the maximum value of the charge on capacitor plates, ω0 is the natural frequencyof the oscillatory circuit, and α is the initial phase. The value of ω0 is determined only by theproperties of the circuit itself, while the values of q0and α depend on the initial conditions. Forthese conditions we can take, for example, the value of the charge q and current i = dq

dt at themoment t = 0.

According to Equation 6, ω0 = 1/√

LC; hence the period of free undamped oscillations isgiven by

T0 = 2π√

LC. (9)

Having found current i by differentiating Equation 8 with respect to time and bearing in mindthat the voltage across the capacitor plates is in phase with charge q, we can easily see that infree undamped oscillations current i leads, in phase, the voltage across the capacitor plates byπ/2.

While solving certain problems, energy approach can also be used.


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1.2.2 Free Damped Oscillations

Every real oscillatory circuit has a resistance, and the energy stored in the circuit is graduallyspent on heating. Free oscillations will be damped. We can obtain the equation for a givenoscillatory circuit by putting V (t) = 0 in Equation 5. This gives

d2q

dt2+ 2γ

dq

dt+ ω2

0q = 0. (10)

I shall give an idea as how to solve this homogeneous differential equation. Form the auxiliaryequation by replacing the differential operator d/dt by x and treating the second derivative asexponentiation:

x2 + 2γx + ω20 = 0 (11)

The solution of this equations are called the eigenvalues of the equation. Once we know theeigenvalues of the homogeneous equation, the most general solution is given as:

q = Aex1t + Bex2t, (12)

where x1, x2 are the eigenvalues; A and B are constants. To get any particular solution, wewill need any two pieces of information to eliminate the constants A and B. This information isgenerally provided in terms of the initial conditions.

Solving (11) for our case, we have the solutions as

x = −γ ±

�γ2 − ω2

0

from where we get the eigenvalues as

x1 = −γ +�

γ2 − ω20, x2 = −γ +

�γ2 − ω2

0. (13)

The following cases arises:

1. γ > ω0: In this case the values of xi’s are real and thus from ( 12) we see that both terms aredecaying exponentials. Hence in this case, the charge q just exponentially falls to zero. Inthis case, the circuit is not oscillatory: it’s over-damped and the discharge of the capacitoris aperiodic.

2. γ = ω0: It is the case of critical damping. We have a repeated solution which is once againnot oscillatory.

3. γ < ω0: This represents under-damped situation. In this case the roots are complex andthe solution turns out to be oscillatory. We study this solution.

For case 3 above, we write the solutions as:

x1,2 = −γ ± ı�

ω20 − γ2 = −γ ± ıωd,

where ı =√−1 and ωd is the damped frequency :

ωd =�

ω20 − γ2 =

�1

LC−

�R

2L

�2

. (14)

After little manipulations, the solution in this case takes the form

q = q0e−γt cos(ωdt + α), (15)


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q0e−γt cos(ωdt)

q0e−γt

q(t)

0 t❘

✾

✻

✲

Figure 3: The variation of the charge q as a function of time for a damped and free oscillations. The graph shownis of the function ( 15) with α set to zero and it shows a cosine wave sandwiched between a decaying exponential.

where q0 and α are arbitrary constants determined from the initial conditions. The plot of thefunction ( 15) is shown in Figure 3. It can be seen that this is not a periodic function since itdetermines damped oscillations. Nevertheless, the quantity T = 2π/ωd is called the period ofdamped oscillations:

T =2π�

ω20 − γ2

=T0�

1− (γ/ω0)2, (16)

where T0 is the period of free undamped oscillations.The factor A(t) = q0e

−γt in ( 15) is called the amplitude of damped oscillations. It’s depen-dence on time is shown in Figure 3 by the dashed line.

Voltage across a Capacitor and Current in an Oscillatory Circuit. Knowing q(t), we canfind the voltage across a capacitor and the current in a circuit. The voltage across a capacitor isgiven by

Vc =q

C=

q0

Ce−γt cos(ωdt + α). (17)

The current in the circuit is

i =dq

dt= q0e

γt[−γ cos(ωdt + α)− ωd sin(ωdt + α)].

We transform the expression in the brackets to cosine. For this purpose, we multiply and dividethis expression by

�ω2

d + γ2 = ω0 and then introduce angle δ by the formulas

− γ

ω0= cos δ,

ωd

ω0= sin δ. (18)

After this, the expression for i becomes:

i = ωdq0e−γt cos(ωdt + α + δ). (19)

It follows from (18) that angle δ lies in the second quadrant (π/2 < δ < π). This means that inthe case of a non–zero resistance, the current in the circuit leads (in phase) the voltage acrossthe capacitor by more than π/2. It should be noted that for R = 0, this lead is δ = π/2.


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1.2.3 Quantities Characterizing Damping

The following quantities characterize any damped circuit.

1. Damping coefficient γ and relaxation time τ , that is the time during which the amplitudeof oscillations decreases by a factor of e. It can be easily seen from ( 15) that

τ =1γ

. (20)

2. Logarithmic decrement λ of damping. It is defined as the Natural log of two successivevalues of the amplitudes measured in a period T of oscillations:

λ = lnA(t)

A(t + T )= γT. (21)

3. Quality factor Q of an oscillatory circuit is, by definition,

Q =π

λ. (22)

where λ is the logarithmic decrement. For the LCR circuit, we can write the quality factoras:

Q =π

γT=

ωd

2γ=

L

R

�1

LC−

�R

2L

�2

. (23)

For a weak damping γ2 � ω20 which means ωd ≈ ω0 the quality factor becomes

Q ≈ ω0L

R. (24)

One more useful form of quality factor for weak damping is:

Q ≈ 2πU

ΔU, (25)

where U is the energy stored in the circuit and ΔU is the decrease in this energy duringthe period T of oscillations.

2 Forced Electric Oscillations. Alternating Current

Steady State Oscillations. We return to the Equations 3 and 5 for an oscillatory circuit andconsider the case when the circuit includes an external voltage which vary with time harmonically:

V (t) = Vm cosωt, (26)

where Vm is the amplitude and ω is the frequency of the voltage. This voltage law occupies aspecial place owing to the properties of the oscillatory circuit itself to retain a harmonic formof oscillations under the action of external harmonic voltage. Further, the standard power pro-duction is of this kind of voltage and current. Before, we take up the solution of the oscillatorycircuit in this case, we first define some terms and introduce some standard notations.


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2.1 Definitions

1. Average or Mean Value: For any continuous function f(x), we define the average valuefav over an interval [a, b] as:

fav =1

b− a

� b

af(x) dx. (27)

Geometrically, fav is the height of the rectangle constructed on the base ab such that itsarea is equal to the area (algebraic) under the graph of the function f(x) bounded by the xaxis, the straight line x = a and x = b (Figure 4). Accordingly, for a harmonically varying

��

��

��

��

fav

a bx

y = f(x)

Figure 4: Geometrical Interpretation of Average value of a function.

voltage or current,Vav = 0.

2. Root Mean Square Value: As the name suggests, the root mean square or RMS is thesquare root of the mean of the squares of the quantities. Accordingly,

frms =

�1

b− a

� b

af2(x) dx. (28)

In particular, for harmonic function V (t) = Vm cos ωt for one complete period T ,

Vrms =

�1T

� T

0V 2

m cos2 ωt d =

�1T

·V 2

m

2

� T

02 cos2 ωt dt

=

�1T

·V 2

m

2

� T

0(1 + cos 2ωt) dt

=

�1T

·V 2

m

2

�t +

sin 2ωt

2ω

�T

0

=

�1T

·V 2

m

2· T =

Vm√2

Hence,

for a harmoic voltage, Vrms =Vm√

2. (29)


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3. Complex Notation. The operator: From Euler’s formula, eıθ = cos θ + ı sin θ. Usingstandard electrical engineering notation, let us denote by =

√−1. Then from Euler’s

formula, we have eθ = cos θ + sin θ. Utilizing this formula, we define a complex voltage:

V = Vm cos ωt + Vm sinωt = Vmeωt, (30)

keeping in mind that the real voltage is the real part of V:

V (t) = Re�V

�. (31)

In this complex notation, the voltage V can be represented as a phasor (a vector)in theComplex plane which rotates with a uniform angular speed of ω. The length/modulus ofthis phasor is the amplitude Vm of the voltage and its instantaneous phase is ωt.

ωt

Vm

V

Vm cos ωtRe

Im

Figure 5: Phasor diagram for harmonic voltage.

4. Reactance and Impedance. In an ac circuit, the capacitors and inductors also poseresistance to the current. The reactance of the capacitor and the inductor is defined as:

Capacitive Reactance: XC =1

ωC, Inductive Reactance: XL = ωL. (32)

Correspondingly, we define the impedances as:

Capacitive Impedance: ZC =1

ωC= −

ωC, Inductive Impedance: ZL = ωL. (33)

The benefit of defining these impedances is that while handling the complex quantities, weadd impedances like resistances and the Ohm’s law ca be used in the same form as wasused for real current; only now the complex quantities replace the real ones.

In polar form, ZC = XCe− π2 and ZL = XCe π

2 . The nice thing about this complexapproach is that impedances can be combined just like resistors.

2.2 Solving an Oscillatory Circuit for Harmonic Input

Figure 6 shows the LCR circuit, where the complex quantities have been indicated. The sourceis a harmonic voltage V (t) = Vm cosωt. We use the method of phasor to solve it. The total


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V = Vmeωt

R XLe π2

XCe− π2

Figure 6: An LCR Circuit

Im

�R2 + (XL −XC)2 e

θ

θ

tan θ = XL−XC

R

ReR

XL = ωL

XC = − 1ωC

(XL −XC)

Figure 7: Phasor representation of the impedancein the complex plane.

impedance of the circuit is:

Z = R + ωL− 1

ωC

= R +

�ωL− 1

ωC

�= R + (XL −XC)

or Z =�

R2 + (XL −XC)2 eθ (34)

where the phase angle θ is given by

tan θ =XL −XC

R. (35)

Applying Ohm’s law, we get the complex current I as:

I =V

Z=

Vmeωt

�R2 + (XL −XC)2 eθ

Thus, we get

I =Vm�

R2 + (XL −XC)2e(ωt−θ) = Im e(ωt−θ), (36)

where we have put

Im =Vm�

R2 + (XL −XC)2, (37)

which is the current amplitude. The real current i(t) in the circuit is just the real part of I,namely:

i(t) = Im cos(ωt− θ). (38)

We study different possible cases for the above circuit.

1. L = 0, C = ∞ (XC = 0), R �= 0 (Purely resistive circuit): In this case, the impedanceof the circuit is just R. Consequently, the phase angle θ = 0 and the current amplitudeIm = Vm/R. The real current is i(t) = (Vm/R) cos ωt. Thus we conclude that in a purelyresistive circuit,

the current and voltage are in phase.


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Im

Re

ωt

V

I

Figure 8: Voltage and current phasor for apurely resistive circuit.

Vm cos ωtIm cos ωt

✛

✁✁

✁✁☛

Figure 9: The variation of voltage and cur-rent in a purely resistive circuit. Ohm’s lawis valid at all instants of time.

The voltage and the current phasor are shown in Figure 8 while the variation of voltageand current as a function of time is shown in Figure 9. Note that in this case, the Ohm’slaw is valid at all instants of time1.

2. L = R = 0 and the capacitance has some finite value (Purely capacitive): In this case,the complex impedance is Z = XCe−π/2 whose modulus is just XC . As such the phaseangle θ = −π/2 and the real current becomes i(t) = ωCVm cos(ωt + π/2) with the resultthat

the current leads the voltage (in phase) by 90◦ in a purely capacitive cir-cuit.

The voltage and the current phasor are shown in Figure 10 while the variation of voltageand current as a function of time is shown in Figure 11. Note that the current amplitudeIm = ωCVm increases as we increase the frequency.

Im

Re

V

ωt

IC

π2

Figure 10: Voltage and current phasor fora purely capacitive circuit.

Vm cos ωt

Im cos(ωt + π/2)

✛

✚✚

✚❃

Figure 11: The variation of voltage andcurrent in a purely capacitive circuit.

3. R = 0, C =∞ (XC = 0), L �= 0 (Purely Inductive Circuit): For this case, the compleximpedance is Z = XLeπ/2 whose modulus is just XL. As such the phase angle θ = π/2and the real current becomes i(t) = (Vm/ωL) cos(ωt− π/2). Conclusion:

the current lags behind the voltage (in phase) by 90◦ in a purely inductivecircuit.

1This assumption holds true only at low frequency. At high frequency, the resistance does depends slightly onthe frequency.


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The voltage and the current phasor are shown in Figure 12 while the variation of voltageand current as a function of time is shown in Figure 13.

Re

Im V

ωt

π2

IL

Figure 12: Voltage and current phasor fora purely inductive circuit.

Vm cos ωt

Im cos(ωt− π/2)

✛

✻

Figure 13: The variation of voltage andcurrent in a purely inductive circuit.

2.3 Power Considerations

The power supplied by the source is given by

P (t) = V (t)i(t) = (Vm cos ωt)(Im cos(ωt− θ)) = VmIm cos ωt cos(ωt− θ). (39)

Using trigonometric formula

2 cos A cos B = cos(A + B) + cos(A−B),

we get from (39),

P (t) =12VmIm[cos(2ωt− θ) + cos θ] =

Vm√2

Im√2[cos(2ωt− θ) + cos θ],

which is the same asP (t) = VrmsIrms[cos(2ωt− θ) + cos θ]. (40)

Thus the power varies as a function of time. The important consideration in an AC circuit is,however, the average power supplied (or absorbed). In the above equation, the first term inthe brackets is a cosine whose average over one complete cycle is zero. Thus the average powersupplied by the source is:

Pav = VrmsIrms cos θ. (41)

In words: The power supplied to the circuit is the product of the rms value of voltageand the rms value of the in phase component of the current.

The term cos θ is called the power factor of the circuit. In terms of the circuit elements,the power factor can be written as

cos θ =R�

R2 + (XL −XC)2. (42)

We note that for a purely resistive circuit, cos θ = 1 and thus the power absorbed by the circuitis VrmsIrms, while for a purely inductive as well as a purely capacitive circuit, cos θ = 0 and thusa purely inductive/capacitive circuit does not absorb any power ! Can you explain this?


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2.4 Back to LCR circuit

Going to the standard solution for the LCR circuit (Figure 6). The solution for the circuit, aswe have already found out, for a sinusoidal voltage V = Vm eωt is given by:

I =Vm�

R2 + (XL −XC)2e(ωt−θ) = Im e(ωt−θ),

wheretan θ =

XL −XC

R.

The complex voltage across the resistor will be, by Ohm’s law:

VR = IR.

Similarly, the complex voltage across the inductor and capacitor are respectively:

VL = IZL, and VC = IZC.

Adding all these complex voltages,

VR + VL + VC = I�R + ZC + ZL

�

=V

R + ZC + ZL

�R + ZC + ZL

�

= V

Thus the sum of the complex voltage drops across the individual elements is the source voltage

V

VC

VC + VL

Re

I

VR

Im

θ

VL

Figure 14: Voltage and current phasor for an LCRcircuit assuming XL > XC .

Vm cos ωt

Im cos(ωt− θ)

✛

✻

Figure 15: The variation of voltage and currentin an LCR circuit assuming XL > XC .

as it must be. This is shown in Figure 14, assuming that XL > XC . Note that in this case,θ > 0 is positive and hence the current lags behind the voltage (if θ had been negative the currentwill lead the voltage). For the same case, the variation of the real voltage and the current as afunction of time is shown in Figure 15.

3 Resonance

Since both XL and XC are functions of the frequency ω of the source, the current amplitude Im

itself is a function of the frequency ω:

Im(ω) =Vm�

R2 +�ωL− 1

ωC

�2. (43)


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From the above expression, we note that as the source frequency is changed, the current am-plitude for the same circuit changes. Specifically, when the frequency is such that XL = XC

the impedance of the circuit is minimum (R) and as such the current amplitude maximum. Atthe same time, the phase angle θ = 0 and the current is in phase with the voltage and thecircuit absorbed maximum power from the source. This phenomena is called resonance. Fromthe condition that XL = XC , we note that it takes place when the frequency of the source isexactly equal to the natural frequency ω0 of the circuit. As such the natural frequency is alsothe resonant frequency:

ωr = ω0 =1√LC

. (44)

The resonance curve for the current amplitude is shown in Figure 16. The maximum currentamplitude at resonance is higher and the sharper, the smaller the damping factor γ = R/2L.

ω0

Im(ω)

ω✲

✻

Figure 16: The variation of the current amplitude as a function of the frequency ω. The different graphscorrespond to different values of the resistance R.

Resonance Curves and Quality factor. The shape of the resonance curve is connected ina certain way with the quality factor of the oscillatory circuit. This connection has the simplestform for the case of weak damping, i.e. for γ2 � ω2

0. In this case, it turns out that

Q =ω0

Δω, (45)

where ω0 is the resonant frequency and Δω is the width of the resonance curve at a “height”equal to 1/

√2 of the peak height, i.e. at resonance.


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