AC Circuit

21.3.7 Resonance in AC circuit

⚪ is defined as the phenomenon that occurs when the
frequency of the applied voltage is equal to the frequency

of the RCL series circuit.

⚪ Figure 21.18 shows the variation of XC, XL, R and Z with
frequency f of the RCL series circuit.

Figure 21.18

51

⚪ From Figure 21.18, the value of impedance is minimum Zmin

when
(21.27

)

where its value is given by

This phenomenon occurs at the frequency fr known as

resonant frequency.

⚪ At resonance in the RCL series circuit, the impedance is

minimum Zmin thus the rms current flows in the circuit is
maximum Imax and is given by

(21.28 52
)

⚪ Figure 21.19 shows the rms current I in RCL series circuit

varies with frequency.

⚪ At frequencies Figure 21.19 resonant frequency fr,
above or below the

the rms current I is less than the rms maximum current Imax

as shown in Figure 21.19.

53

⚪ The resonant frequency, fr of the RCL series circuit is given by

and

(21.29
)

⚪ The series reswohnearnece circuit is used for tuning a radio
receiver.

Note:

At resonance, the current I and voltage V are in phase.

54

Example 21.5 :

A 2 μF capacitor and a 1000 Ω resistor are placed in series with an
alternating voltage source of 12 V and frequency of 50 Hz.
Calculate
a. the current flowing,
b. the voltage across the capacitor,
c. the phase angle of the circuit.
Solution :
a. The capacitive reactance of the inductor is given by

and the impedance of the circuit is

55

Solution :
a. Therefore the current flowing in the circuit is
b. The voltage across the capacitor is given by
c. The phase angle between the current and supply voltage is

OR
56

Example 21.6 :

Figure 21.20

Based on the RCL series circuit in Figure 21.20 , the rms voltages

across R, L and C are shown.

a. With the aid of the phasor diagram, determine the applied voltage

and the phase angle of the circuit.

Calculate:

b. the current flows in the circuit if the resistance of the resistor R is

26 Ω,

c. the inductance and capacitance if the frequency of the AC source

is 50 Hz,

d. the resonant frequency.

57

Solution :
a. The phasor diagram of the circuit is

From the phasor diagram,

the applied voltage V is

and the phase angle φ is

OR 58

Solution :
b. Given

Since R, C and L are connected in series, hence the current

passes through each devices is the same. Therefore

c. Given
The inductive reactance is

thus the inductance of the inductor is

59

Solution :
c. Meanwhile, the capacitive reactance is

thus the capacitance of the capacitor is
d. The resonant frequency is given by

60

Exercise 21.1 :

1. An AC current of angular frequency of 1.0 × 104 rad s−1 flows
through a 10 kΩ resistor and a 0.10 μF capacitor which are

connected in series. Calculate the rms voltage across the

capacitor if the rms voltage across the resistor is 20 V.

ANS. : 2.0 V
2. A 200 Ω resistor, a 0.75 H inductor and a capacitor of

capacitance C are connected in series to an alternating

source 250 V, 600 Hz. Calculate

a. the inductive reactance and capacitive reactance when

resonance is occurred.

b. the capacitance C.

c. the impedance of the circuit at resonance.
d. the current flows through the circuit at resonance. Sketch

the phasor diagram of the circuit.
ANS. : 2.83 kΩ, 2.83 kΩ; 93.8 nF; 200 Ω; 1.25 A

61

3. A capacitor of capacitance C, a coil of inductance L, a resistor
of resistance R and a lamp of negligible resistance are placed
in series with alternating voltage V. Its frequency f is varied
from a low to a high value while the magnitude of V is kept

constant.
a. Describe and explain how the brightness of the lamp varies.

b. If V=0.01 V, C =0.4 μF, L =0.4 H, R = 10 Ω and the

circuit at resonance, calculate
i. the resonant frequency,
ii. the maximum rms current,
iii. the voltage across the capacitor.

(Advanced Level Physics,7th edition, Nelkon & Parker, Q2, p.423)

ANS. : 400 Hz; 0.001 A; 1 V

62

www.kmph.matrik.edu.my Learning Outcome:

21.4 Power and power factor (1 hour)

At the end of this chapter, students should be able to:
⚪ Apply

● average power,

● instantaneous power,

● power factor,

in AC circuit consisting of R, RC, RL and RCL in series.

63

21.4 Power and power factor

21.4.1 Power of a pure resistor

⚪ In a pure resistor, the voltage V and current I are in phase,
thus the instantaneous power P is given by

and

(21.30

where )

⚪ Figure 21.21 shows a graph of instantaneous power P being

absorbed by the resistor against time t.

64

Power being absorbed

⚪ The average (or meFaing)uproew2e1r.P21av being absorbed by the

resistor is given by

(21.31
)

65

21.4.2 Power of a pure capacitor
⚪ In a pure capacitor, the current I leads the voltage V by π/2

radians, thus the instantaneous power P is given by

and
(21.32
)

⚪ Figure 21.22 shows a graph of instantaneous power P of the
pure capacitor against time t.

66

Power being absorbed

Power being returned to

⚪ The avesraugpepl(yor meFaing)uproew2e1r.P22av of the pure capacitor is given

by

67

21.4.3 Power of a pure inductor
⚪ In a pure inductor, the voltage V leads the current I by π/2

radians, thus the instantaneous power P is given by

and

⚪ Figure 21.23 shows a graph of instantaneous power P of the
pure inductor against time t.

68

Power being absorbed

Power being returned to

⚪ The avesraugpepl(yor meFaing)uproew2e1r.P23av of the pure inductor is given

by
Note:
The term ‘resistance’ is not used in pure capacitor and inductor because no
heat is dissipated from both devices.

69

21.4.4 Power and power factor of R, RC, RL and
RCL series circuits

⚪ In an AC circuit in which there is a resistor R, an inductor L and
a capacitor C, the average power Pav is equal to that dissipated

from the resistor i.e.

(21.33
)

rms values

⚪ From the phasor diagram of the RCL series circuit as shown in

Figure 21.24,

Figure 21.24

70

We get

then the eq. (21.33 ) can be written as
and

(21.34

where cos φ is called the power factor of the AC )circuit, Pr is
the average real power and I2Z is called the apparent power.

⚪ Power factor is defined as

(21.35
)

Note: where

From the Figure 21.24, the power factor also can be calculated by using the
equation below:

(21.36

) 71

Example 21.7 :

A 100 μF capacitor, a 4.0 H inductor and a 35 Ω resistor are
connected in series with an alternating source given by the
equation below:

Calculate:
a. the frequency of the source,
b. the capacitive reactance and inductive reactance,
c. the impedance of the circuit,
d. the peak current in the circuit,
e. the phase angle,
f. the power factor of the circuit.

72

Solution :

By comparing to the

Thus

a. The frequency of AC source is given by

b. The capacitive reactance is

and the inductive reactance is

73

Solution :
c. The impedance of the circuit is

d. The peak current in the circuit is

74

Solution :
e. The phase angle between the current and the supply voltage is

OR
f. The power factor of the circuit is given by

75

Example 21.8 :

A 22.5 mH inductor, a 105 Ω resistor and a 32.3 μF capacitor are
connected in series to the alternating source 240 V, 50 Hz.
a. Sketch the phasor diagram for the circuit.
b. Calculate the power factor of the circuit.
c. Determine the average power consumed by the circuit.
Solution :

a. The capacitive reactance is

and the inductive reactance is

76

Solution :
a. Thus the phasor diagram for the circuit is

b. From the phasor diagram in (a),
the impedance of the circuit is

77

Solution :
b. and the power factor of the circuit is
c. The average power consumed by the circuit is given by

and

78

Exercise 21.2 :

1. An RLC circuit has a resistance of 105 Ω, an inductance of

85.0 mH and a capacitance of 13.2 μF.

a. What is the power factor of the circuit if it is connected to a
125 Hz AC generator?

b. Will the power factor increase, decrease or stay the same
if the resistance is increased? Explain.

(Physics, 3rd edition, James S. Walker, Q47, p.834)

ANS. : 0.962; U think
2. A 1.15 kΩ resistor and a 505 mH inductor are connected in

series to a 14.2 V,1250 Hz AC generator.

a. What is the rms current in the circuit?
b. What is the capacitance’s value must be inserted in series

with the resistor and inductor to reduce the rms current to
half of the value in part (a)?

(Physics, 3rd edition, James S. Walker, Q69, p.835)

ANS. : 3.44 mA, 10.5 nF 79