Alternating Current, Class 12, Chapter Notes, Physics (IIT - JEE & AIPMT)

Alternating Current and its Average and RMS Value - Alternating Current, Class 12, Physics

Alternating Current

1. Alternating Current

Until now, we have studied only circuits with direct current (dc) which flows only in one direction (as shown in fiugre I and II). The primary source of emf in such circuit is a battery. When a resistance is connected across the terminals of the battery, a current is established in the circuits, which flows in a unique direction from the positive terminal to the negative terminal via the external resistance.

But most of the electric power generated and used in the world is in the form ofalternating current (ac), the magnitude of which changes continuously with time and direction is reversed periodically (as shown in figure III & IV)and it is given by

i = i₀ sin (ωt + φ)

Here i is instantaneous value of current i.e., the magnitude of current at any instant of time and i₀ is the maximum value of current which is called peak current or the current amplitude and the current repeats its value after each time interval T =  as shown in figure. This time interval is called the time period and w is angular frequency which is equal to 2π times of frequency f.

ω = 2πf

The current is positive for half the time period and negative for remaining half period. It means that the direction of current is reversed after each half time period. The frequency of ac in India is 50 Hz.

An alternating voltage is given by

V = V₀ sin (ωt + φ)

It also varies alternatively as shown in the figure (b), where V is instantaneous voltage and V₀ is peak voltage. It is produced by ac generator also called as ac dynamo.

AC Circuit : An ac circuit consists of circuit element i.e., resistor, capacitor, inductor or any combination of these and a generator that provides the alternating current as shown in figure. The ac source is represented by symbol  in the circuit.

2. AVERAGE AND RMS VALUE OF ALTERNATING CURRENT

2.1 Average current (Mean current)

As we know an alternating current is given by

i = i₀ sin (ωt + f) ...(1)

The mean or the average value of ac over any time T is given by

Using equation (1)

In one complete cycle, the average current

 = - =  = 0 (as ωT = 2π)

Since ac is positive during the first half cycle and negative during the other half cycle so i_avg will be zero for long time also. Hence the dc instrument will indicate zero deflection when connected to a branch carrying ac current. So it is defined for either positive half cycle or negative half cycle.

 ₌  0.637 i₀

Similarly V_avg =  0.637 V₀

Ex.1 If a direct current of value a ampere is superimposed on an alternating current i = b sin wt flowing through a wire, what is the effective value of the resulting current in the circuit ?

Sol. As current at any instant in the circuit will be,

i = i_dc + i_ac = a + b sin ωt

So, i_eff =  = 

i.e., = 

but as

 = 0 and 

So, i_eff = 

2.2 R.M.S Value of alternating current

The notation rms refers to root mean square, which is given by square root of mean of square current.

i.e., 

i²_avg =  

=  = 

=  = 

 = 

i_rms =  » 0.707 i₀

Similarly the rms voltage is given by

V_rms =  » 0.707 V₀

The significance of rms current and rms voltage may be shown by considering a resistance R carrying a current i = i₀ sin (wt + f)

The voltage across the resistor will be

V_R = Ri = (i₀R) sin (ωt + )

The thermal energy developed in the resistor during the time t to t + dt is

i² R dt = i₀²R sin²(ωt + ) dt

The thermal energy developed in one time period is

U =  =  = RT  = i²_rms RT

It means the root mean square value of ac is that value of steady current, which would generated the same amount of heat in a given resistance in a given time.

So in ac circuits, current and ac voltage are measured in terms of their rms values. Likes when we say that the house hold supply is 220 V ac it means the rms value is 220 V and peak value is = 311 V.

Ex.2 If the voltage in an ac circuit is represented by the equation, V = sin (314t - ),calculate (a) peak and rms value of the voltage, (b) average voltage, (c) frequency of ac.

Sol. (a) For ac voltage,

V = V₀ sin (ωt -  )

The peak value of voltage

V₀ =  = 311 V

The rms value of voltage

(b) Average voltage in full cycle is zero. Average voltage in half cycle is

V_avg =  =  = 198.17 V

(c) As ω = 2πf, 2ωf = 314

i.e., f =  = 50 Hz

Ex.3 The electric current in a circuit is given by i = i₀ (t/T) for some time. Calculate the rms current for the period t = 0 to t = T.

Sol. The mean square current is

 = 

Thus, the rms current is

i_rms = 

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Series AC Circuit - Alternating Current, Class 12, Physics

3. SERIES AC CIRCUIT

3.1 When only Resistance is in an AC circuit

Consider a simple ac circuit consisting of a resistor of resistance R and an ac generator, as shown in the figure. According to Kirchhoff's loop law at any instant, the algebraic sum of the potential difference around a closed loop in a circuit must be zero.

ε - V_R = 0

ε - i_RR = 0

ε₀ sinωt - i_RR = 0

i_R =  sinωt = i₀ sin ωt ..(i)

where i₀ is the maximum current. i₀ = 

From above equations, we see that the instantaneous voltage drop across the resistor

V_R = i₀R sinωt ...(ii)

We see in equation (i) & (ii), i_R and V_R both vary as sin wt and reach their maximum values at the same time as shown in figure (a), they are said to be in phase. A phasor diagram is used to represent phase relationships. The lengths of the arrows correspond to V₀ and i₀. The projections of the arrows onto the vertical axis give V_R and i_R. In case of the single-loop resistive circuit, the current and voltage phasors lie along the same line, as shown in figure (b), because i_R and V_R are in phase.

3.2 When only Inductor is in An AC circuit

Now consider an ac circuit consisting only of an Inductor of inductance L connected to the terminals of an ac generator, as shown in the figure. The induced emf across the inductor is given by Ldi/dt. On applying Kirchhoff's loop rule to the circuit

ε- V_L = 0 ⇒ ε - L

When we rearrange this equation and substitute

ε = ε₀ sin ωt, we get   

 = ε₀ sin ωt ...(iii)

Integration of this expression gives the current as a function of time

i_L =  = - 

For average value of current over one time period to be zero, C = 0

Therefore, i_L = - 

When we use the trigonometric identity

coswt = - sin(wt - p/2), we can express equation as

i_L =  ...(iv)

From equation (iv), we see that the current reaches its maximum values when cos wt = 1.

i₀ =  =  ...(v)

where the quantity X_L, called the inductive reactance, is

X_L = ωL

The expression for the rms current is similar to equation (v), with ε₀ replaced by ε_rms.

Inductive reactance, like resistance, has unit of ohm.

We can think of equation (v) as Ohm's law for an inductive circuit.

On comparing result of equation (iv) with equation (iii), we can see that the current and voltage are out of phase with each other by π/2 rad, or 90º. A plot of voltage and current versus time is given in figure (a). The voltage reaches its maximum value one quarter of an oscillation period before the current reaches its maximum value. The corresponding phasor diagram for this circuit is shown in figure (b). Thus, we see that for a sinusoidal applied voltage, the current in an inductor always lags behind the voltage across the inductor by 90º.

Ex.4 An inductor of inductance L = 5 H is connected to an

AC source having voltage v = 10 sin (10t + )

Find 

(i) Inductive Reactance (x_L)

(ii) Peak & Rms voltage (V₀ & V_rms)

(iii) Peak & Rms current (I₀ & I_rms)

(iv) Intstantatious current (I_(t))

Sol. (i) x_L = ωL = 10 × 5 = 50

(ii) v₀ = 10

v_rms = 

(iii) 

I_rms = 

(iv) I(t) = 

3.3 When only Capacitor is in An AC circuit

Figure shows an ac circuit consisting of a capacitor of capacitance C connected across the terminals of an ac generator. On applying Kirchhoff's loop rule to this circuit, we get

ε - V_C = 0

V_C = ε = ε₀ sin ωt ...(vi)

where V_C is the instantaneous voltage drop across the capacitor. From the definition of capacitance, V_C = Q/C, and this value for V_C substituted into equation gives

Q = C ε₀ sin ωt

Since i = dQ/dt, on differentiating above equation gives the instantaneous current in the circuit.

Here again we see that the current is not in phase with the voltage drop across the capacitor, given by equation (vi). Using the trigonometric identity cos ωt = sin(ωt + π/2), we can express this equation in the alternative from

 ...(vii)

From equation (vii), we see that the current in the circuit reaches its maximum value when cos ωt= 1.

Where X_C is called the capacitive reactance.

The SI unit of X_C is also ohm. The rms current is given by an expression similar to equation with V₀ replaced by V_rms.

Combining equation (vi) & (vii), we can express the instantaneous voltage drop across the capacitor as

V_C = V₀ sin ωt = i₀ X_C sin ωt

Comparing the result of equation (v) with equation (vi), we see that the current is π/2 rad = 90º out of phase with the voltage across the capacitor. A plot of current and voltage versus time, shows that the current reaches its maximum value one quarter of a cycle sooner than the voltage reaches its maximum value. The corresponding phasor diagram is shown in the figure (b). Thus we see that for a sinusoidally applied emf, the current always leads the voltage across a capacitor by 90º.

Brain Teaser

What is the reactance of a capacitor connected to a constant DC source ?

Ex.5 A capacitor of capacitive reactance 5? is connected with A.C. source having voltage V = 3 sin (ωt + p/6). Find rms and Peak voltage rms and peak current and instantaneous current

Sol. On comparing with

 ⇒ v₀ = 3

 ⇒ 

 ⇒ I(t) = I₀ sin 

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Series L-R, C-R and C-R Circuit - Alternating Current, Class 12, Physics

4. SERIES L-R CIRCUIT

Now consider an ac circuit consisting of a resistor of resistance R and an inductor of inductance L in series with an ac source generator.

Suppose in phasor diagram, current is taken along positive x-direction. The V_R is also along positive x-direction and V_L along positive y-direction as we know that potential difference across a resistance in ac is in phase with current and it leads in phase by 90º with current across the inductor, and as we know V_R = i₀R & V₀ = i₀X_L

i = i₀ sin wt  

V_R(t) = i₀ Rsin ωt

V_L(t) = i₀ X_L sin (ωt + p/2)

hence we can write

V(t) = i₀R sin ωt + i₀ X_L sin (ωt + p/2)

V₀ = i₀

where  is known as impedence (z) of the circuit.

now we can write

where tan β= 

hence β = 

Ex.6 When 100 volt dc is applied across a coil, a current of 1 amp flows through it; when 100 V ac of 50 Hz is applied to the same coil, only 0.5 amp flows. Calculate the resistance of inductance of the coil.

Sol. In case of a coil, i.e., L - R circuit.

i =  with Z =  = 

So when dc is applied, ω = 0, so z = R

and hence i =  i.e., R =  =  = 200 ?

i.e., ω²L² = Z²-R²

i.e., (2πfL)² = 200² - 100² = 3 × 10⁴ (as ω = 2πf)

So, L =  =  = 0.55 H

Ex.7 A 12 ohm resistance and an inductance of 0.05/p henry with negligible resistance are connected in series. Across the end of this circuit is connected a 130 volt alternating voltage of frequency 50 cycles/second. Calculate the alternating current in the circuit and potential difference across the resistance and that across the inductance.

Sol. The impedance of the circuit is given by

Z =  = 

=  =  = 13 ohm

Current in the circuit i = E/Z =  = 10 amp

Potential difference across resistance

V_R = iR = 10 × 12 = 120 volt

Inductive reactance of coil X_L = ωL = 2πfL

Therefore, X_L = 2π × 50 ×  = 5 ohm

Potential difference across inductance

V_L = i × X_L = 10 × 5 = 50 volt

5. SERIES C-R CIRCUIT

Now consider an ac circuit consisting of a resistor of resistance R and an capacitor of capacitance C in series with an ac source generator.

Suppose in phasor diagram current is taken along positive x-direction. Then V_R is also along positive x-direction but V_C is along negative y-direction as potential difference across a capacitor in ac lags in phase by 90º with the current in the circuit. So we can write,

V_R = I₀ R sin ωt 

Potential difference across capacitor

Potential at any instant t

V(t) = V₀ sin (ωt + b)

tan α = 

Ex.8 An A.C. source of angular frequency w is fed across a resistor R and a capacitor C in series. The current registered is i. If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency.

Sol. At angular frequency w, the current in R-C circuit is given by

i_rms =  ...(i)

When frequency is changed to ω/3, the current is halved. Thus

 =  ...(ii)

...(ii)

From equations (i) and (ii), we have

 = 

Solving this equation, we get 

Hence, the ratio of reactance to resistance is 

Ex.9 A 50 W, 100 V lamp is to be connected to an ac mains of 200 V, 50 Hz. What capacitance is essential to be put in series with the lamp?

Sol. As resistance of the lamp R =  = 200 ? and the maximum current i =  =  =  ; so when the lamp is put in series with a capacitance and run at 200 V ac, from V =iZ we have,

Z =  =  = 400?

Now as in case of C-R circuit, Z = ,

i.e., R² +  = 160000

or,  = 16 × 10⁴ - (200)² = 12 × 10⁴

So,  =  × 10²

or C = 

i.e., C =  = 9.2 μF

5. L.C. circuit

As shown in figure a capacitor and inductance are connected in series method and alternating voltage is applied across the circuit.

Let X_c is capacitance reactance,

X_L is Inductance reactance,

i = i₀ sin wt current floing through the circuit

V_C(t) = i₀ X_C sin (ωt - π/2)

V_L(t) = i₀ X_L sin (ωt + π/2)

= i₀ X_C sin wt cos π/2 - i₀ X_C cos wt sin π/2 + i₀ X_L sin wt cos π/2 + i₀ X_L cos wt sin π/2

= i₀ cos w t(X_L - X_C)

V(t) = V₀sin (wt + p/2)

V₀ = i₀Z 

Z = (X_L - X_C)

cos = 0

V_CO = i₀ X_C ; 

6. SERIES L-C-R CIRCUIT

Now consider an ac circuit consisting of a resistor of resistance R, a capacitor of capacitance C and an inductor of inductance L are in series with an ac source generator.

Suppose in a phasor diagram current is taken along positive x-direction. Then V_R is along positive x-direction, V_L along positive y-direction and V_C along negative y-direction, as potential difference across an inductor leads the current by 90° in phase while that across a capacitor, lags by 90°.

V = 

L - R - C circuit

Impedance phasor of above circuit

& Impedance triangle

here B is phase angle By triangle tan β= 

Power factor cos=  

Let I be the current in the series circuit of any instant then

(1) Voltage V(t) = V₀ sin (ωt + β) = i₀ z sin (ωt + β)

here v₀ = i₀z & v_rms = i_rmsz

(2)  here voltage V_L across the inducetance is ahead of current I in phase by π/2 rad

(3) V_C(t) = V_{O C} sin (ωt - π/2)

here voltage V_C across the capacitance lags behind the current I in pase by π/2 rad

(4) V_R(t) = i₀ R sin ωt

here voltage V_R across the resistor R has same phase as I

V_{O R} = I_O R

Special Case :

(1) When X_L > X_C or V_L > V_C then emf is ahead of current by phase β which is given by

tan β =  or cos f = 

The series LCR circuit is said to be inductive

(2) When X_L < X_C or V_L < V_C then current is ahead of emf by phase angle β which is given by

 or

The series LCR circuit is said to be capacitive

(3) When X_L = X_C or V_L = V_C, b = 0, the emf and current will be in the same phase. The series LCR circuit is said to be purely resistive. It may malso be noted that

 or  or I_Rms = 

Susceptance : The reciprocal of the reactane of an a.c. circuit is called its susceptance.

Admittance : The reciprocal of the impedance of an a.c. circuit is called its admittance.

Ex.10 Figure shows a series LCR cicuit connected to a variable voltage source V = 10 sin (ωt + p/4) ;

x_L = 10 ?, X_C = 6 ?, R = 3 ?

Calculate Z, i₀, i_rms, v_rms, V_{L O}, V_{C O}, V_{R O}, b,

V_{L Rms}, V_{C Rms,} V_Rms, i(t), V_L(t), V_c(t), and V_R(t)

X_L > X_C

Sol. V = 10 sin (wt + p/4) so V₀ = 10 volt

V_rms = 

Therefore, Z = 

; 

; 

; 

i(t) = 20 sin (ωt + π/4 - 53°)

V_L(t) = 20 sin (ωt + π/4 - 53° + π/2)

= 20 sin (ωt +  - 53° - ) = 12 sin (ωt -  - 53°)

Ex.11 A resistor of resistance R, an inductor of inductance L and a capacitor of capacitance C all are connected in series with an a.c. supply. The resistance of R is 16 ohm and for a given frequency the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find

(a) the potential difference across R, L and C (b) the impedance of the circuit

(c) the voltage of a.c. supply (d) phase angle

Sol. (a) Potential difference across resistance

V_R = iR = 5 × 16 = 80 volt

Potential difference across inductance

V_L = i × (wL) = 5 × 24 = 120 volt

Potential difference across condenser

V_C = i × (1/wC) = 5 × 12 = 60 volt

(b)  ==20 ohm

(c) The voltage of a.c. supply is given by

V = iZ = 5 × 20 = 100 volt

(d) Phase angle

=  = 

= tan^-1(0.75) = 36°87'

Ex.12 A series circuit consists of a resistance of 15 ohm, an inductance of 0.08 henry and a condenser of capacity 30 micro farad. The applied voltage has a frequency of 500 radian/s. Does the current lead or lag the applied voltage and by what angle?

Sol. Here ωL = 500 × 0.08 = 40 ohm

and 

= 66.7 ohm

tan=  = - 1.78

= - 60.67°

Thus the current leads the applied voltage by 60.67°

Ex.13 A current of 4 A flows in a coil when connected to a 12 V d.c. source. If the same coil is connected to a 12 V, 50 rad/s, a.c. source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a 2500 μF condenser is connected in series with the coil.

Sol. When the coil is connected to a d.c. source, its resistance R is given by

R =  =  = 3 ?

When it is connected to a.c. source, the impedance Z of the coil is given by

Z =  =  = 5?

For a coil, Z = 

Therefore, 5 = 

or 25 = [(3)² + (50L)²]

Solving we get L = 0.08 henry

When the coil is connected with a condenser in series, the impedance Z' is given by

Z' =  = 

= 5 ohm

Power developed P = V_rms × i_rms × cosf

where cos = R/Z' = 3/5 = 0.6

Therefore, P = 12 × 2.4 × 0.6 = 17.28 watt

Ex.14 A 20 volts 5 watt lamp is used in ac main of 220 volts 50 c.p.s. Calculate the (i) Capacitance of capacitor. (ii) Inductance of inductor, to be put in series to run the lamp, (iii) What pure resistance should be included in place of the above device so that the lamp can run on its voltage. (iv) Which of the above arrangements will be more economical and why ?

Sol. The current required by the lamp

The resistance of the lamp

So for proper running of the lamp, the current through the lamp should be 0.25 amp

(i) When the condenser C is placed in series with lamp, then

The current through the circuit

 or 

Solving it for C, we get

C = 4.0 × 10^-6 F = 4.0 μF

(ii) When inductor L henry is placed in series with the lamp, then

  or  

or 

Solving it for L, we get L = 2.53 henry

(iii) When resistance r ohm is placed in series with lamp of resistance R, then

or 

(iv) It will be more economical to use inductance or capacitance is series with the lamp to run it as it consumes no power while there would be dissipation of power when resistance is inserted in series with the lamp.

Brain Teaser

Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit ?

Power in AC Circuit - Alternating Current, Class 12, Physics

7. Power in an AC circuit

In case of a steady current the rate of doing work is given by,

P = Vi

In an alternating circuit, current and voltage both vary with time, so the work done by the source in time interval dt is given by

dW= Vidt

Suppose in an ac, the current is leading the voltage by an angle. Then we can write,

V = V₀ sinωt

and i = i₀ sin(ωt +)

dW = V₀i₀ sin ωt sin (ωt + ) dt

= V₀ i₀ (sin² ωt cos f + sinωt cos ωt sin) dt

The total work done in a complete cycle is

W = V₀i₀ cos + V₀i₀sin

+  = 

The average power delivered by the source is, therefore,

P =  =  = 

= V_rms i_rms cos

or <P>_{one cycle} = V_rms i_rms cos

Here, the term cos  is known as power factor.

It is said to be leading if current leads voltage, lagging if current lags voltage. Thus, a power factor of 0.5 lagging means current lags the voltage by 60° (as cos^-10.5 = 60°). The product of V_rms and i_rms gives the apparent power. While the true power is obtained by multiplying the apparent power by the power factor cos. Thus,

and apparent power = V_rms × i_rms

True power = apparent power × power factor

For = 0°, the current and voltage are in phase. The power is thus, maximum (V_rms× i_rms). For

= 90°, the power is zero. The current is then stated as wattless. Such a case will arise when resistance in the circuit is zero. The circuit is purely inductive or capacitive.

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Resonance - Alternating Current, Class 12, Physics

8. Resonant Frequency

A series LCR circuit is said to be in the resonance condition when the current through it has its maximum value.

The current amplitude I₀ for a series LCR circuit is given by

Clearly I₀ becomes zero both for ω → 0 and ω → ∞. The value of I₀ is maximum when

 or 

⇒ 

Then impedance will be minimum

Z_min = R

The circuit is purely resistive. The current and voltage are in the same phase and the current in the circuit is maximum. This condition of the LCR circuit is called resonance condition. 

The variance of I₀ v/s ω shown in following figure

So cos =  =  = 1

V = V₀ sin (ωt)

Impedance phase of resonance circuit

Impedance of the circuit is minimum and heat generated in the circuit is maximum.

Ex.15 In following LCR circuit find Z, i(t), V_OC, V_OL at resonace frequency

Sol. Z = Z_min = R = 2?

V_{O L} = i₀X_C = 100 volt

V_{O L} = i₀ X_L = 100 volt

 : Above circuit is used as voltage amplifier (magnification) as peak value of voltage by source is only 10 while we can have maximum voltage up to 100 (V_{O C}& V_{O L})

Ex.16 A series LCR with R = 20 ?, L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V a.c. supply. When the frequency of the supply equals the natural frequency of the circuit. What is the average power transferred to the circuit in one complete cycle?

Sol. When the frequency of the supply equals the natural frequency of the circuit, resonance occurs.

Therefore, Z = R = 20 ohm

i_rms = 

Average power transferred/cycle

P = E_rmsi_rms cos0° = 200 × 10 × 1 = 2000 watt

8.1 Sharpness of Resonance (Q - factor) :

The Q- factor of a series resonant circuit is defined as the ratio of the resonant frequency to the difference in two frequencies taken on the both sides of the resonant frequency such that at each frequency, the current amplitude becomes times the value of resonant frequency.

Mathematically Q-factor.

or 

9. Choke Coil :

A choke coil is simply an inductor with large inductance which is used to reduce current in a.c. circuit without much loss of energy.

Principle. A choke coil is based upon the principle that when a.c. flows through an inductor, the current lags behind the e.m.f. by a phase angle π/2.

Construction. A choke coil is basically an inductance. It consists of a large number of turns of insulated copper wire wound over a soft iron core. In order to minimise loss of electrical energy due to production of eddy currents, a laminated iron core is used.

In practice, a low frequency choke coil is made of insulated copper wire wound on a soft iron core, while a high frequnecy choke coil has air as core materials

Working : As shown in fig a choke is put in series across an electrical appliances of resistance R and is connected to an a.c. source.

Average power dissipiated per cycle in the circuit is

P_av = V_eff I_eff cosf = V_eff I_eff  .

Inductance L of the choke coil is very large so that R << wL. Then

Power factor cos   0   

tan= 

Uses. In a.c. circuit, a choke coil is used to control the current in place of a resistance. If a resistance is used to control the current, the electrical energy will be wasted in the form of heat. A choke coil decreases the current without wasting electrical energy in the form of heat.

10. OSCILLATIONS IN L-C CIRCUIT

If a charged capacitor C is short-circuited through an inductor L, the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is dissipated as heat. Assume an ideal situation in which energy is not radiated away from the circuit. With these idealizations-zero resistance and no radiation, the oscillations in the circuit persist indefinitely and the energy is transferred from capacitor's electric field to the inductor's magnetic field back and forth. The total energy associated with the circuit is constant. This is analogous to the transfer of energy in an oscillating mechanical system from potential energy to kinetic energy and back, with constant total energy. Such an analogous mechanical system is an example of spring mass system.

Let us now derive an equation for the oscillations of charge and current in an L-C circuit. Refer figure (a) : The capacitor is charged to a potential difference V such that charge on capacitor q₀ = CV

Here q₀ is the maximum charge on the capacitor. At time t = 0, it is connected to an inductor through a switch S. At time t = 0, the switch S is closed.

Refer figure (b) : When the switch is closed, the capacitor starts discharging. Let at time t charge on the capacitor is q (<q₀) and since, it is further decreasing, there is a current i in the circuit in the direction shown in figure.

The potential difference across capacitor = potential difference across inductor, or

V_b - V_a = V_c - V_d

Therefore,  ...(1)

Now, as the charge is decreasing, i = 

or  = - 

Substituting in equation (1), we get

or  = -  ...(2)

This is the standard equation of simple harmonic motion 

Here w =  ...(3)

The general solution of equation (2), is

q = q₀ cos (ωt ± ) ...(4)

In our case = 0 as q = q₀ at t = 0.

Thus, we can say that the charge in the circuit oscillates with angular frequency ω given by equation (3). Thus,

ln L - C oscillations, q, i and  all oscillate simple harmonically with same angular frequency ω, but the phase difference between q and i or between i and  is . Their amplitudes are q₀ q₀ω are ω² q₀ respectively. So

q = q₀ cosωt, then ...(5)

i = - = q₀ω sin ωt ...(6)

and cosωt ...(7)

Potential energy in the capacitor

U_C =  =  ...(8)

Potential energy in the inductor

U_L =  =  ...(9)

Thus potential energy stored in the capacitor and that in the inductor also oscillates between maximum value and zero with double the frequency. All these quantities are shown in the figures that follows

Ex.17 A capacitor of capacitance 25 μF is charged to 300 v. It is then connected across a 10 μH inductor. The resistance of the circuit is negligible.

(a) Find the frequency of oscillation of the circuit.

(b) Find the potential difference across capacitor and magnitude of circuit current 1.2 ms after the inductor and capacitor are connected.

(c) Find the magnetic energy and electric energy at t = 0 and t = 1.2 ms.

Sol. (a) The frequency of oscillation of the circuit is ,

f = 

Substituting the given values we have,  = 

(b) Charge across the capacitor at time t will be ,

q = q₀ cos ωt

and i = - q₀ ωsin ωt

Here q₀ = CV₀ = (25 × 10^-6) (300) = 7.5 × 10^-3 C

Now, charge is the capacitor after t = 1.2 × 10^-3 s is,

q = (7.5 × 10^-3) cos (2p × 318.3) (1.2 × 10^-3)C

= 5.53 × 10^-3C

Therefore, P.D. across capacitor,

V =  = 221.2 volt

The magnitude of current in the circuit at

t = 1.2 × 10^-3 s is,

|i| = q₀ ω sinωt

= (7.5 × 10^-3) (2p) (318.3) sin(2p × 318.3) (1.2 × 10^-3) A = 10.13 A

(c) At t = 0 : Current in the circuit is zero. Hence,U_L = 0

Charge on the capacitor is maximum

Hence, U_c = 

or U_c =  = 1.125 J

Therefore, Total energy E = U_L + U_C = 1.125 J

At t = 1.2 ms

U_L =  = (10.13)² = 0.513 J

U_C = E - U_L = 1.125 - 0.513 = 0.612 J

Otherwise U_C can be calculated as,

U_C =  =  = 0.612 J

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Transformers - Alternating Current, Class 12, Physics

11. Transformer :

A transformer is an electrical device for converting an alternating current at low voltage into that at high voltage or vice versa. If it increase the input voltage, it is called step up transformer and if it decreases the input voltage, it is called step down transformer.

Principle : It works on the principle of mutual induction, i.e., when a changing current is passed through one of the two inductively coupled coils, an induced emf is setup in the other coil.

Construction : A transformer essentially consists of two coils of insulated copper wire having different number of turns and wound on the same soft iron core. The coil to which electric energy is supplied is called the primary and the coil from which energy is drawn or output is obtained is called the secondary.

To prevent energy losses due to eddy currents, a laminated sheet is used. Because of high permeability of soft iron, the entire magnetic flux due to the current in the primary coil practically remains in the iron core and hence passes fully through the secondary.

Two types of arrangements are generally used for winding of primary and secondary coils in a transformer.

1. Core type : In this type, the primary and secondary coils are wound on seperate limbs of the core.

2. Shell type : In this type the primary and secondary coils are wound one over another on the same limb of the iron core.

Theory : Consider the situation when no load is connected to the secondary, i.e., its terminals are open. Let N₁ and N₂ be the number of terminal in the primary and secondary respectively. Then Induced emf in the primary coil

Induced emf in the secondary coil

whereis the magnetic flux linked with each turn of the primary or secondary at any instant. Thus

The ratio  of the number of turns in the secondary to that in the primary called the turns ratio of the transformer. It is also called transformation ratio

for step up transformer : N₂ > N₁

for step down transformer : N₁ > N₂

Currents in primary and secondary : Assuming the transformer to be ideal one so that there are no energy loss, then

Input power = output power

or V_p I_p = V_s I_s

where I_p and I_S are the currents in the primary and secondary respectively.

Hence 

Efficiency : The efficiency of a transformer is defined as

η =  × 100%

The efficiency of real transformer is fairely high (90 - 98%) though not 100%

Extra Portion For IIT-Mains

Explain with the help of a labelled diagram, the principle and working of an a.c. generator ? Write the expression for the emf generated in the coil in terms of speed of rotation. Can the current produced by an a.c. generator be measured with a moving coil galvanometer.

Sol. AC generator : A dynamo or generator is a device which converts mechancial energy into electrical energy. It is based on the principle of electromagnetic induction.

Construction : It consists of the four main parts :

(i) Field Magnet : It produces the magnetic field. In the case of a low power dynamo, the magnetic field is generated by a permanent magnet, while in the case of large power dynamo, the magnetic field is produced by an electromagnet.

(ii) Armature : It consists of a large number of turns of insultated wire in the soft iron drum or ring. It can revolve reound an axle between the two poles of the field magnet. The drum or ring serves the two purposes : (i) It serves as a support to coils and (ii) It increases the magnetic field due to air core being replaced by an iron core :

(iii) Slip Rings : The slip rings R₁ and R₂ are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature.

(iv) Brushes : These are two flexible metal plates or carbon rods (B₁ and B₂) which are are fixed and constantly touch the revolving rings. The ouput current in extyernal load R_L is taken through these brushes.

Working : When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming's right hand rule. Considering the armature to be in vertical positoin and as it rotates

in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B₁ B_L B₂. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit if flows along B₂ R_L B₁. Thus the direction of induced emf and current changes in the external circuit after each half revolution.

If N is the number of turns in coil, f the frequency of rotation. A area of coil and B the magnetic induction, then induced emf

Obviously, the emf produced is alternating and hence the current is also alternating

Current produced by an ac generator cannot be measured by moving coil ammeter, because the average value of ac over fully cycle is zero.