A.C CIRCUIT CONTAINING INDUCTANCE
When alternating voltage is applied across a pure inductive coil a back (E) is induced in the coil due to its self inductance.
E = –
The negative sign indicates that induced e. m .f opposes the change in current.
In order to maintain the flow of current the applied voltage must be equal and opposite to induced voltage at every instant.
Consider a pure inductor of inductance L connected across an alternating source of
E = Eosin ωt
Suppose the instantaneous value of the alternating e. m. f is given by
Esin wt ………. ( )
If I is the current in the circuit and is the rate of change of current at that instant, then e. m. f induced in L is given by
As applied voltage is equal and opposite to induce e .
m .f at every instant
From, Equation (i) and Equation (ii)
Integrating both sides we get
The value of I will be maximum Io when, Sin = 1
Then,
Substituting the value of = in equation (iii)
I = sin (wt –
……………….. (iv)
I) Phase Angle
It is clear from equation (i ) and (iv) that circuit current lags behind the applied voltage by (/2) radians or 900.
This fact is also indicated in the wave diagram.
The phase diagram in figure below also reveals the fact that lags behind by 90º

Hence in an a.c circuit current through L lags behind the voltage across L by 90º
This means that when voltage across L is zero, current through L is maximum and vice versa.
From
E = L
Now, is maximum when circuit current is zero and is zero when circuit current is maximum.

INDUCTIVE REACTANCE
Inductive reactance is the opposition in which an inductor offers to current flow. It is denoted by XL
Inductance not only causes the current to lag behind the voltage but it also limits the magnitude of current in the circuit.
We have seen above that,
Io =
ωL =
Clearly the opposition of inductance to current flow is ωL. This quantity ωL is called inductive reactance XL of the inductor.
Inductive reactance XL
a) From
XL =
But = and Io =
Then,

XL =

b) For d.c
f = 0
so that,
XL = 2
XL = 2
Therefore a pure inductance offers zero opposition to d.c

c) XL = 2
Therefore,the greater f,the greater is XL and vice versa.

d) We can show that the units of XL are that of ohm
XL = ωL = x Henry = and
XL =

III) Average Power consumed
E = sin ωt
I = sin (ωt- I = – cos ωt
Instantaneous power P
P = EI
P = (sin ωt) (-cos ωt)
P = – sin ωt cos ωt
P = Sin 2ωt
Average power P is equal to average of power over one cycle.
P = Sin 2ωtdt

P = 0
Hence average power absorbed by pure inductor is zero
During one quarter cycle of alternating source of e .m .f. energy is stored in the magnetic field of the inductor this energy is supplied by the source.
During the next quarter cycle the stored energy is returned to the source. For this reason average power absorbed by a pure inductor over a complete cycle is zero.

NUMERICAL EXAMPLES
1. A pure inductive coil allows a current of 10A to flow from a voltage of 230V and frequency 60Hz supply.
Find
i) Inductive reactance
ii) Inductance of the coil
iii) Power consumed
Write down equations for voltage and current:

Solution
EV = 230V
IV = 10A
f =50Hz
i) Inductive reactance XL
XL = =
ii) Inductance of the coil L
From
XL =2
L = 0.073 H
iii) Power absorbed = 0
Also
= 230 x = 10 x
= 325.27V = 14.14A
ω = 2
ω = 314
Since in pure Inductive circuit current lags behind the applied voltage by radians. The equation for voltage and current are,
E = 325 .27 sin 314t, I = 14.14 sin (314t)
2. Calculate the frequency at which the inductive reactance of 0.7H inductor is 220Ω
Solution
XL = 220 Ω
L = 0.7H
f =?
f =
f = 50 Hz

3. A coil has self inductance of 1.4H. The current through the coil varies sinusoidally with amplitude of 2A and frequency 50 Hz
Calculate
i) Potential difference across the coil
ii) r .m .s value of P.d across the coil.

Solution
(i) P.d across the coil
E = L
E = L
E = L
E = L cos t,
E = L 2 cos2
E = 2
E = 880 cos 100
ii) r.m.s value of potential different across the coil
=
=
= 622.2V

4. How much inductance should be connected to 200V, 50 Hz a.c supply so that a maximum current of 0.9A flows through it?
Solution
= 200V
Io = 0.9A
f = 50 Hz
Peak value of voltage
=
=
Inductive reactance L
XL =
XL =
XL = 314.27 Ω
Inductance L,
L =
L=
L = 1 H

5. An Inductor of 2H and negligible resistance is connected to 12V, 50Hz supply. Find the circuit current, what current flows when the inductance is changed to 6H?

Solution
* For the First case XL
XL = 2
XL = 2
XL = 628 Ω
Circuit current
=
=
= 0.019A
* For the second case XL
XL = 2
XL = 2
XL = 1884 Ω
Circuit current
=
= 0.0063A

A.C CIRCUIT CONTAINING CAPACITANCE ONLY
When an alternating voltage is applied to a capacitor, the capacitor is charged first in one direction and then in the opposite direction.

The result is that electrons move to and fro round the circuit connecting the plates, thus constituting alternating current.
Consider a capacitor of capacitance C connected across an alternating source of e .m. f .
Suppose the instantaneous value of the alternating e.m.f E is given by
E = sin ωt …….. (i)
If I is the current in the circuit and Q is the charge on the capacitor at this instant, then the Potential difference across the capacitor VC
At every instant the applied e.m.f E must be equal to the potential difference across the capacitor.
E =
Q = C sin ωt
From,
I =
I =
I = Cωcos ωt
I = Cω sin (ωt + 2

The value of I0 will be maximum Io when sin (wt+) = 1
Substituting the value of Io in equation (ii)

1) Phase Angle
It is clear from equation (I) and (ii) that circuit current leads the applied voltage by π/2 radians or 90Ëš.
This fact is also indicated in the wave diagram.

It also reveals that Iv leads Ev by 90°, hence in a.c circuit current in capacity leads the voltage.
This means that when voltage across capacitor is zero, current in capacitor is maximum and vice versa.
When P.d across capacitor is maximum, the capacitor is fully charged, i.e circuit current is zero.
Since the rate at which a sinusoidally varying p.d falls is greater as it reaches zero, the current has its maximum value when P.d across capacitor is zero. Hence, current and voltage are out of phase by 90°.

CAPACITIVE RESISTANCE
Capacitive resistance is the opposition which a capacitor offers to current flow. It is denoted by XC.
Capacitance not only causes the voltage to lag behind the current but it also limits the magnitude of current in the circuit.
We have seen above that
=
=

Then
Clearly, the opposition offered by capacitance to current flow is 1/ωC
The quantity 1/ωC is called capacity reactance XC of the capacitor
XC will be in Ω if f is in Hz and C in Farad
a) From
=
Then
=

b) For d.c
f = 0
XC = =
XC =

XC = ∞
Therefore a pure capacitance offers infinite opposition to d.c. In other words, a capacitor blocks d.c.
c. From
XC =

Therefore the greater the f the smaller is XC and vice versa
d) From
Therefore the units XC are Ohm

(iii) AVERAGE POWER ABSORBED
From
E = sin ωt
I = sin (ωt+2)
I = cos ωt
Instantaneous power P
P = EI
P = Sin ωt.
Cos ωt
P = (sinωtcosωt)
P = Sin 2ωt
Average power P = Average of P Over one cycle
P =Sin 2ωtdt

P = 0
Hence average power absorbed by pure capacitance is zero.
During one quarter cycle of the alternating source of energy is stored in the electric field of the capacitor.
This energy is supplied by the source during the next quarter cycle, the stored energy is returned to the source

WORKED EXAMPLES
1. A 318μF capacitor is connected 230V, 50Hz supply. Determine
i) The capacitive reactance
ii) r.m.s value of circuit current
iii) Equations for voltage and current
Solution
C = 318μF = 318 x 10-6 F
f = 50 Hz
i) Capacitive reactance XC
XC =
XC =
XC = 10 Ω
ii) r.m.s value of current
= =
= 23A
iii ) = . = √2 x 23
= 230 = 32.53A
= 325.27 V
ω= 2
ω = 2
ω= 314
E0 = 325.27sin 314t and I = 32.53 sin314t

2. A coil has an inductance of 1H
a) At what frequency will it have a reactance of 3142Ω?
b) What should be the capacitance of a condenser which has the same reactance at that frequency?

Solution
a) L = 1H
XL = 3142Ω
f = ?
XL = 2
f = = :
f= 500Hz
b) XC = 3142 , f=500 Hz
C =?
XC =
C =
C =
C = 0.11
C = 0.11
3. A 50 capacitor is connected to a 230V, 50Hz supply. Determine
i) The maximum charge on the capacitor
ii) The maximum energy stored in the capacitor

Solution
The charge and energy in capacitor will be maximum when p.d across the capacitor is maximum.
i) Maximum charge on the capacitor
Q = C
Q = C
Q = (50x) x (230 x)
Q = 16.26 x C
ii) Maximum energy stored in the capacitor U
U = C
U = 1/2 x (50x) x (230 x 2
U = 2.65J
4. The Instantaneous current in a pure inductance of 5H is given be
I = 10sin (314t- amperes
A capacitor is connected in parallel with the inductor. What should be the capacitance of the capacitor to receive the same amount of energy as inductance at the same terminal voltage?

Solution
The current flowing through pure inductor is
I = 10 Sin (314t)
= 10A
ω = 314s-1
Maximum energy stored in the inductor
UL = 250J ………. (i)
Now
= ωL
= 314 x 5 x 10
= 15700v
Max energy stored in the capacitor of capacitance C
=
= ……….(ii)
Equate the equation (I and (ii)
= = 250
C =
C = 2.03 x F
C = 2.03

A.C CIRCUIT CONTAINING R AND L IN SERIES
Consider a resistor of resistance R ohms connected in series with pure inductor of L Henry.

Let
= r.m.s value of applied alternating e.m.f
=r.m.s value of the circuit current
VR = R when VR is in phase with
VL = where leads by 90º
Taking current as the reference phasor, the phasor diagram of the circuit can be drawn as shown in figure.

The voltage drop VR is in phases with current and is represented in magnitude and direction by the phase OA.
The voltage drop VL leads the current by 90º and is represented in magnitude and direction by the phase AB.
The applied voltage is the phasor sums of these two voltage drops
= +
=
=
=

1) Phase Angle
It is clear from the phasor diagram that circuit current lags behind the applied voltage by Φº.
Therefore we arrive at a very important conclusion that in an inductive circuit current lags behind the voltage.

NUMERICAL EXAMPLE
1. Three impedance are connected in series across a 200V, 50Hz supply. The first impedance is a 10â„¦ resistor and the second is a coil or 15 â„¦ inductive reactance and 5 â„¦ resistance while the third consists of a 15 â„¦ resistor in series with a 25â„¦ capacitor
Calculate
i) Circuit current
ii) Circuit phase angle
iii) Circuit power factor
iv) Power consumed

Solution
i) Total circuit Resistance
R = 10 +5 + 15
R = 30 â„¦
Total Circuit reactance
X = XL – XC
X = 15 – 25
X = -10 â„¦ (capacitive)
Circuit impedance Z
Z = 31. 6 â„¦
Circuit current IV
IV = E v
Z
I V = 200
31.6
IV = 6. 33 A

ii) Circuit phase Angle
iii) Circuit power factor
Power factor = COS θ
= COS 18 .26º
Power factor = 0. 949
iv) Power consumed P
P = EV IV Cos
P = (200 x 6.33) x 0. 949
P = 1201. 4 W
Alt
P = Iv2 R

2. A 230V, 50Hz supply is applied to a coil of 0.06 H inductance and 2.5 Ω resistance connected in series with 6.8 μF capacitor
Calculate
i. Circuit impedance
ii. Circuit current
iii. Phase angle between EV and Iv
iv. Power factor
v. Power consumed

Solution
i) Inductive reactance XL
XL = 2ΠfL
XL = 2Πx 50 x 0.60
XL = 18 .85 Ω
Capacitive reactance XC
XC = 468 Ω
Circuit Impedance
Z = 449. 2 Ω

ii) Circuit current IV
IV = 0.152 A
iii) Phase angle between EV and IV
iv) Power factor
Cos Φ = 0.0056
v) Power consumed P
P = EVIS Cos Φ
P = 230.x 0.512 x 0.0056
P = 0.66 W
3. A resistance R, and inductance L = 0.01H and a capacitance C are connected in series. When an alternating voltage E = 400Sin (3000t- 200) is applied to the series combination, the current flowing is 10√2 Sin (3000t – 650). Find the value of R and C

Solution
The circuit current lags behind the applied voltage by θ
θ = 650 – 20
θ = 450
This implies that the circuit is inductive i.e.
XL > XC
The net circuit reactance X
X = XL –XC
Now
XL = ωL
XL = 3000 x 0.01
XL = 30 Ω
Also
From
Circuit Impedance Z
Z = 28.3 Ω
Now
Z2 = R2 + X2
Z = R2 + R2
Z2 = 2R2
Z = R √2
R = 20Ω
Now
X = XL – XC
20 = 30 – XC
XC= 10Ω
From

C = 33.3 x 10 -6 F
4. A series RLC circuit is connected to an a.c (220V, 50 H) as shown in the figure below

If the reading of the three volt meter V1 V 2 and V3 are 65V, 415V and 204V respectively.
Calculate
i. The current in the circuit
ii. The value of inductor L
iii. The value of capacitor C

Solution

Here voltmeters are considered ideal i.e. having infinite resistance.
Therefore, it is a series RLC circuit
i) Circuit current IV
IV = 0. 65 A
ii) Inductive reactance XL
XL = 318. 85 Ω
Inductance L
L = 1H
iii) Capacitive reactance XC
XC = 638. 46Ω
Capacitance C

C = 5 x 10-6 F

5. A coil of resistance 8Ω and inductance 0.03H is connected to an a.c supply of 240V, 50 Hz.
Calculate
i) The current the power and power factor.
ii) The value of a capacitance which when connected in series with the above coil causes no change in the value of current and power taken from the supply

Solution
i) Reactance of the coil XL
XL=9.42 Ω
Impedance of the coil Z
Z = 12. 46 Ω
Circuit Current IV
IV = 19.42 A
Power consumed

P= (19.42)2 x 8
P= 3017 W
Power factor Cos Φ
Cos Φ = 0.65 lag

ii) To maintain the same current and power, the impedance of the circuit should remain unchanged. Thus the value of capacitance in the series circuit should be such so as to cause the current to lead by the same angles as it previously lagged.
This can be achieved if the series capacitor has a capacitive reactance equal to twice the inductive reactance.
XC = 2XL
XC = 2 x 9 .42
XC = 18.84 Φ
Now
C = 169 x 10-9 F

RESONANCE IN R – L – C SERIES A.C CIRCUIT

The R- L – C series A.C circuit is said to be in electrical resonance when the circuit power factor is unity
XL = XC
This is called series resonance.

Resonant frequency
This is the frequency at which the reactance of the coil has the same magnitude and so giving the current in a circuit its maximum value.
XL = XC
where
XC = capacitive reactance
XL= inductive reactance
The frequency at which resonance occurs is called the resonant frequency
The resonance in R – L – C series circuit can be achieved by changing the supply frequency because XL and XC are frequency dependent
At a certain frequency, called the resonant frequency fo, XL becomes equal to XC and resonance occurs.
At resonance
XL = XC
If L and C are in Henry and farad respectively, then ƒo will be in Hz

EFFECTS OF SERIES RESONANCE

When series resonance occurs, the effect on the circuit is the same as though neither inductance nor capacitance is present.

The current under this condition is dependent solely on the resistance of the circuit and voltage across it.
(i)The impedance of the circuit is minimum and equal to the resistance of the circuit.
From
When Z=ZR, XL = XC

ZR = R
At series resonance
(ii) The circuit current is maximum as it is limited by the resistance of the circuit alone.
(iii) Since at series resonance the current flowing in the circuit is very large, the voltage drops across L and C are also very large.
In fact, these drops are much greater than the applied voltage.
However, voltage drop across L – C combination as a whole will be zero because these drops are equal in magnitude but 1800 out of phase with each other.

RESONANCE CURVE

Resonance curve is the curve between the circuit current and the supply frequency.
Figure below shows the resonance curve of a typical R-L-C series circuit.

Current reaches the maximum value at the resonant frequency fo, falling off rapidly on either side at that point.

It is because if the frequency is below fo, XC > XL and the net reactance is no longer zero.

Net reactance X

X = XC – XL

If the frequency is above fo, then XL > XC and the net reactance is again not zero.
Net reactance X

X = XL – XC
In both cases, the circuit impedance will be more than the impedance ZR at resonance.
The result is that magnitude of circuit current decreases rapidly as the frequency changes from the resonant frequency.
The effect of resistance in the circuit. The smaller the resistance, the greater is the current at resonance and sharper the resonance curve.
On other hand, the greater the resistance, the lower is the resonant peak.

Q – FACTOR OF SERIES RESONANT CIRCUIT

At series resonance, the p.d across L or C (the two voltage drop being equal and opposite) builds up to a value many times greater than the applied voltage EV.
The voltage magnification produced by series resonance is termed as Q – Factor of the series resonant circuit.
Q – Factor of a resonant.

R-L circuit
This is the ration of voltage across L or C to the applied voltage.
Or
This is the ratio of power stored to power dissipated in the circuit reactance and resistance respectively.
Where
X = Capacitive or inductive reactance at resonance
R = Series resistance
The Q – factor of a series resonant circuit can also be expressed in terms of L and C.
From,

Also

The value of Q-factor depends entirely upon the design of the coil i.e. R – L part of the R – L – C circuit because resistance arises in this rather than.
With a well designed coil, the quality factor can be 200 or more.

PHYSICAL MEANING OF Q – FACTOR

The Q- Factor of series a.c circuit indicates how many times the p.d across L or C is greater than the applied voltage at resonance.
For example, consider on R-L- C series circuit connected to 240v a.c source.
If Q – factor of the coil is 20, then voltage across L or C will be,

VC = VL = QVR = 20 x 240
VC = VL = 4800 V

Q – FACTOR AND RESONANCE CURVE

At series resonance the circuit current is maximum (IR = EV/R) and is limited by circuit resistance only.
The smaller the circuit resistance, the greater is the circuit current and sharper will be the resonance curve.

Smaller circuit resistance means large value of Q – Factor. Therefore, the greater the Q – Factor of resonant R-L-C circuit, the sharper is the resonance curve.

APPLICATION OF SERIES RESONANCE

One important applicant of series resonance is to tune radio and TV receivers.

The input signal comes from the antenna and induces a voltage E in L of the series resonant circuit.

VC = QE
The voltage across the capacitor becomes VC.
where:
Q = Quality factor of the circuit
As the value of Q is generally large, the original signal received by the antenna increases many times in value and appears across C.
The value of VC is much more than that could have been obtained by direct transformer ratio.
Thus amplifier receives a greatly increased signal.

Bandwidth of a series resonant circuit.
This is the range of frequencies over which circuit current is equal to or greater than 70.7% of maximum current. (IR, current at resonance).
The two frequencies are cut off frequency f1 and upper off frequency f2.
Bandwidth (BW) = f2 – f1
It can be shown by using mathematical equation:-
The circuit has a general appearance of a parallel circuit but actually it is a series circuit.

It is because no separate volt is applied to L but instead a voltage E is induced in it which is co as a voltage in series with LC.

Q – Factor of a resonant Circuit
Is the ration of resonant frequency fo to the bandwidth of the circuit

RESONANCE IN PARALLEL A.C CIRCUIT
A parallel a.c circuit containing reactive element (L and C) is said to be in electrical resonance when the circuit power factor is unity. This is called parallel resonance
Consider a pure inductor of inductance L connected in parallel with a capacitor of capacitance across an a .c source of voltage EV ( r. m . s)

The circuit will be in resonance when the circuit power factor is unity.
This means that wattles component of the circuit current should be zero.
IL – IC=0
The resonance in a parallel a .c circuit can be achieved by changing the supply frequency because XL and XC are frequency dependent .
At a certain frequency called resonant frequency f0, IC become equal to IL and resonance occur.

At resonance
I C = IL
From

f0 will be in Hz if L is in Henry and C is in farad

EFFECTS OF PARALLEL RESONANCE
1) The circuit power factor becomes unity. This implies that the circuit act as a resistor

2) The impedance (resistive) of the circuit becomes maximum.

3) The circuit current is minimum the small current IV flowing in the circuit is only the amount needed to supply the resistance losses.

RESONANCE CURVE
This is the curve between the circuit current and the supply frequency.

Figure below shows the resonance curve of a parallel a.c circuit.

The circuit current IV is minimum at parallel resonance.
As the frequency changes from resonance, the circuit current increases rapidly.
This action can be explained as follows. For frequencies other than the resonance, the reactive currents (IL and IC) in the two branches of the circuit are not equal.
The resultant reactive current must be supplied by the a.c source .
As the difference of the reactive currents in the two branches increase with the amount of deviation from the resonant frequency, the circuits current will also increase.

(i)Alternating voltages can be stepped up or stepped down efficiently by a transformer.
This permits the transmission of electric power at high voltages to achieve economy and distribute the power at utilization voltages.

(ii)A.C motors are cheaper and simpler in construction than the d.c motors.

(iii) A. C can be easily converted into d.c by rectifiers.

(iv)Alternating current can be controlled with a choke coil without any appreciable loss of electrical energy.

(v) The switch gear (e.g. switches, circuit breakers) for a.c system is cheaper than the d.c system.

(i) For the same voltage (same value of voltage), a.c is more dangerous than d.c.

(ii) The shock of a.c is attractive whereas that of d.c is repulsive.

(iii) A.C cannot be used for some processes e.g. electroplating, charging of batteries etc.

(iv) A.C is transmitted more over the surface of the conductor than from inside. This is called skin effect.

To avoid skin effect, a.c is transmitted over several fine insulated wires instead of a single thick wire.

WORKED EXAMPLES
1. A coil of resistance 100Ω and inductance 100μH is connected in series with a 100 pF capacitor. The circuit is connected to a 10V variable frequency source.

Calculate:
(i) Resonant Frequency
(ii) Current at resonance
(iii) Voltage across L and C at resonance.

Solution:
Capacitance C = 100pF = 100 x 10-12 F
Inductance L = 100μH = 100 x 10-6 H
(i) Resonant frequency f0

f0 = 1 . 59 x 106 HZ

(ii) Current at resonance IR

IR = 0.1 A

(iii) Voltage across L

VL = Ir XL

VL = 100V

(iv) Voltage across C

VC = 100V
At series resonance , the voltage across L or C is much greater than the applied voltage.

2. A circuit, having a resistance of 4Ω and inductance of 0.5 H and a variable capacitance in series, is connected across a 100V, 50 Hz supply. Calculate
(i) The capacitance to give resonance
(ii) The voltage across inductance and capacitance.

Solution
(i) At series resonance
XL = XC
C = 20.26 x 10-6F
(ii) Current at resonance Ir.
Ir = 25 A
P.d across L
VL = Ir XL
VL = 3927V
P.d across C
VC = Ir XC
VC = 3927 V

3. A series R – L- C circuit consists of a 100 Ω resistor, an inductor of 0.318 H and a capacitor of unknown value. When this circuit is energized by √2 x 230 sin 314t volts are supply the current is found to be √2 x2.3 sin 314t find.

(i) The value of capacitor in microfarad.
(ii) Voltage across indicator
(iii) Total power consumed.

Solution
From,

EO = √2 EV
E = √2 x 230 Sin 314t
E=Ev Sin ωt
Ev = √2 x 230
√2 x230 = √2 x EV

EV = 230V.
Circuit current IV
Iv = 2.3 A
Also further phase angle is zero
ω = 314 S-1
f = 50 Hz
i) At resistance
XL = XC
C = 31.4 x 10 -6 F
C = 31.4 x μF
ii) Voltage across inductor VL
VL = 230 V

iii) Total power consumed
P= 2.32 x 100
P = 529 W

4. A coil of inductance 8 μH is connected to a capacitor of capacitor of capacitance 0.02 μF. To what wavelength is this circuit is tuned?

Solution
L = 8 x 10 -6 H
C = 0.02 x 10 -6F
Resonant frequency f0
f0 = 3.98 x 105 Hz
If C = 3. 108 m /s is the velocity of the e.m wave, then wavelength λ

λ= 7.54 X 102 m

5. a ) A Sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to A series L.C. R circuit in which R = 3 Ω , L = 25.48 μH and C = 796 μF . Find
i) The impendence of the circuit.
ii) The phase difference between voltage across the source and the current
iii) The power dissipated in the circuit
iv) Power factor.
b) Suppose the frequency of the source in the previous example can be varied.
i) What is the frequency of the source at which resonance occurs?
ii) Calculate the impedance the current and the power dissipate at the resonant condition.

Solution
a) E0 = 283 V
f= 50 Hz
R = 3Ω
L = 25.48 x 10– 3 H
C = 796 x 10-6 F
XL = 8Ω
XC = 4Ω

i) Circuit impedance Z
Z = √25
Z = 5 Ω

ii) The phase difference between voltage and circuit current

iii) Circuit current IV
IV = 40 A

iv) Power dissipated in the circuit
P = IV R
P = 402 x 3
P = 4800 W
Power factor = Cos 53.1º
b) i) the frequency at which resonance occur

f0 = 35.4 HZ

ii) The impedance z at resonance is equal to R
Z = 3Ω
The circuit Current IV
IV = 66 .7 A
Power dissipate at resonance is
P = (66.7) x 3
P = 13350 W
L – C IN PARALLEL CIRCUIT.
Consider inductor of inductance L Henry connected in parallel with capacitor of capacitance C.

Figure below shows the two component of current.

IL lags by 90º on EV but IC leads by 90º on E

CASE 1
If IC is greater than IL at the particular frequency
Since I leads by 90º on E in this case we say that the circuit in net capacitive.

CASE 2
If IL is greater than IC
I = IL – IC
Since I lags by 90º on E in this case, the circuit is Net inductive.

CASE 3
At the Resonance
XC = XL

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