Chapter 7 Alternating Current

NCERT Solutions Class 12 Physics Chapter 7 - Alternating Current

The physics chapter for Class 12 on alternating current (AC) follows the work done in previous chapters on electromagnetic induction and expands into the concept of alternating current (AC). The chapter discusses how AC can theoretically be defined and gives students a mathematical understanding of AC through angle (or phasors) and how AC takes place in the electrical components of a circuit.

To develop a strong understanding of this chapter, it is necessary for students to understand Electrical Engineering (also called Modern Electricity), Resonance (Communication Systems), and Electrical Distribution (Power Transmission). Therefore, this chapter will be an important one for the CBSE Boards, JEE, and NEET. Additionally, the NCERT solutions for Chapter 7 provide students with many examples that demonstrate the phase relationships between voltage and current using both analytical methods and phasor diagrams. The early examples provided by these solutions are based on much simpler circuits, whereas the later examples are based on more complex series of LCR circuits.

1.0Class 12 Physics Chapter 7 Alternating Current: Key Concepts

Class 12 Physics Chapter 7, Alternating Current, explains the behaviour of electrical circuits when the voltage and current change with time. The main concepts covered in this chapter include:

RMS and Peak Values: Understand the relationship between the peak value and RMS (effective) value of alternating current and voltage.
AC Through Resistor, Inductor and Capacitor: Learn how alternating current behaves in a resistor, inductor, and capacitor. In a resistor, current and voltage are in phase. In an inductor, current lags behind voltage by one-quarter cycle, while in a capacitor, current leads voltage by one-quarter cycle.
Series LCR Circuit: Study the combined behaviour of a resistor, inductor, and capacitor connected in series and learn how to calculate the total impedance of the circuit.
Resonance and Q-Factor: Understand resonance in an LCR circuit, where impedance is minimum and current is maximum. The chapter also explains the sharpness of resonance using the Q-factor.
Power in AC Circuits: Learn about power consumption in AC circuits, power factor, and wattless current.
LC Oscillations: Understand how energy is exchanged between an inductor and a capacitor in an LC circuit.
Transformers: Study how transformers use mutual induction to increase or decrease AC voltage and learn about the main causes of energy loss in transformers.

2.0NCERT Solutions Class 12 Physics Chapter 7 Alternating Current : Detailed Solutions

SOLVED EXAMPLES

A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source and (c) the rms current through the bulb Sol. (a) We are given $P = 100 W$ and $V = 220 V$ . The resistance of the bulb is

$R = \frac{V ^{2}}{P} = \frac{( 220 V ) ^{2}}{100 W} = 484Ω$

(b) The peak voltage of the source is

$V_{m} = 2 V = 100 W$

$I = \frac{P}{V} = \frac{100 W}{220 V} = 0.454 A$

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz . Sol. The inductive reactance,

$X_{L} = 2 π vL = 2 \times 3.14 \times 50 \times 25 \times 1 0^{- 3} Ω = 7.85Ω$

The rms current in the circuit is

$I = \frac{V}{X _{L}} = \frac{220 V}{7.85Ω} = 28 A$

A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Sol. When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitive reactance $(1/ ω C)$ and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.
A $15.0 μ F$ capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Sol. The capacitive reactance is

$X_{C} = \frac{1}{2 π ν C} = \frac{1}{2 \times 3.14 \times ( 50 Hz ) ( 15.0 \times 1 0 ^{- 6} F )} = 212Ω$

The rms current is $I = \frac{V}{X _{C}} = \frac{220 V}{212Ω} = 1.04 A$ The peak current is

$i_{m} = 2 I = (1.41) (1.04 A) = 1.47 A$

This current oscillates between +1.47A and -1.47 A , and is ahead of the voltage by $π /2$ . If the frequency is doubled, the capacitive reactance is halved and consequently, the current is double.

A light bulb and an open coil inductor are connected to an ac source through a key as shown in figure.

Circuit diagram showing a light bulb and open-coil inductor connected to an AC source through a key.

The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron is inserted. Give your answer with reasons. Sol. As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

A resistor of $200Ω$ and a capacitor of $15.0 μ F$ are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Sol. Given : $R = 200Ω, C = 15.0 μ F = 15.0 \times 1 0^{- 6} F$

$V = 220 V, v = 50 Hz$

(a) In order to calculate the current, we need the impedance of the circuit. It is

$Z = R^{2} + X_{C}^{2} = R^{2} + (2 π ν C)^{2} = (200Ω)^{2} + (2 \times 3.14 \times 50 \times 15.0 \times 1 0^{- 6} F)^{2} = (200Ω)^{2} + (212.3Ω)^{2} = 291.67Ω$

Therefore, the current in the circuit is

$I = \frac{V}{Z} = \frac{220 V}{291.5Ω} = 0.755 A$

(b) Since the current is the same throughout the circuit, we have

$V_{R} = I R = (0.755 A) (200Ω) = 151 V V_{C} = I X_{C} = (0.755 A) (212.3Ω) = 160.3 V$

The algebraic sum of the two voltages, $V_{R}$ and $V_{C}$ is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem :

$V_{R + C} = V_{R}^{2} + V_{C}^{2} V_{R + C} = V_{R}^{2} + V_{C}^{2} = 220 V$

Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

(a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

Sol.

(a) We know that $P = IV cos ϕ$ where $cos ϕ$ is the power factor. To supply a given power at a given voltage, if $cos ϕ$ is small, we have to increase current accordingly. But this will lead to large power loss $(I^{2} R)$ in transmission. (b) Suppose in circuit, current I lags the voltage by an angle $ϕ$ . Then power factor $cos ϕ = R / Z$ . We can improve the power factor (tending to 1 ) by making Z tend to R. Let us understand, with the help of a phasor diagram (figure)

Phasor diagram showing voltage V, current I, in-phase current Ip, quadrature current Iq, and phase angle φ.

how this can be achieved. Let us resolve I into two components. $I_{p}$ along the applied voltage $V$ and $I_{q}$ perpendicular to the applied voltage. $I_{q}$ is called the wattless component since corresponding to this component of current, there is no power loss. $I_{p}$ is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It's clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current $I_{q}$ by an equal leading wattless current $I_{q}^{'}$ . This can be done by connecting a capacitor of appropriate value in parallel so that $I_{q}$ and $I_{q}^{'}$ cancel each other and P is effectively $I_{p} V$ .

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which $R = 3Ω. L = 25.48 mH$ and $C = 796 μ F$ . Find (a) the impedance of the circuit ; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.

Sol.

(a) To find the impedance of the circuit, we first calculate $X_{L}$ and $X_{C}$ .

$X_{L} X_{C} = 2 π ν L = 2 \times 3.14 \times 50 \times 25.48 \times 1 0^{- 3} Ω = 8Ω = \frac{1}{2 π ν C} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 1 0 ^{- 6}} = 4Ω$

Therefore,

$Z = R^{2} + (X_{L} - X_{C})^{2} = 3^{2} + (8 - 4)^{2} = 5Ω$

(b) Phase difference,

$ϕ = tan^{- 1} \frac{X _{C} - X _{L}}{R} = tan^{- 1} (\frac{4 - 8}{3}) = - 53. 1^{\circ}$

Since $ϕ$ is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is $= I^{2} R$ Now,

$I = \frac{i _{m}}{2} = \frac{1}{2} (\frac{283}{5}) = 40 A$

Therefore, $P = (40 A)^{2} \times 3Ω = 4800 W$ (d) Power factor $= cos ϕ = cos (- 53. 1^{\circ}) = 0.6$

Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.

Sol.

(a) The frequency at which the resonance occurs is

$ω_{0} = \frac{1}{LC} = \frac{1}{25.48 \times 1 0 ^{- 3} \times 796 \times 1 0 ^{- 6}} = 222.1 rad / s v_{r} = \frac{ω _{0}}{2 π} = \frac{221.1}{2 \times 3.14} Hz = 35.4 Hz$

(b) The impedance Z at resonant condition is equal to the resistance :

$Z = R = 3Ω$

The rms current at resonance is,

$I = \frac{V}{Z} = \frac{V}{R} = (\frac{283}{2}) \times \frac{1}{3} = 66.7 A$

The power dissipated at resonance,

$P = I^{2} \times R = (66.7)^{2} \times 3 = 13.35 kW$

You can see that in the present case, power dissipated at resonance is more than the power dissipated in Question-8.

At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Sol. The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes - resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

EXERCISE QUESTIONS WITH SOLUTIONS

A $100Ω$ resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?

Sol. $R = 100Ω, V_{rms} = 220 V, f = 50 Hz$

(a) The rms value of current in the circuit,

$I_{rms} = \frac{V _{rms}}{R} = \frac{220}{100} = 2.20 A$

(b) The net power consumed over a full cycle,

$P_{avg} = V_{rms} I_{rms} cos ϕ = V_{rms} I_{rms} cos 0^{\circ} = 220 \times 2.2 \times 1 = 484 W$

(a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? Sol. (a) $V_{0} = 300 V$ rms voltage,

$V_{rms} = \frac{V _{0}}{2} = \frac{300}{2} = 212.16 V$

(b) The rms value of current, $I_{rms} = 10 A$ Now peak current,

$I_{0} = I_{rms} 2 = 102 = 14.1 A$

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. Sol. $L = 44 mH = 44 \times 1 0^{- 3} H$ ,

$V_{rms} = 220 V, v = 50 Hz$

Angular frequency, $ω = 2 π v$ Inductive reactance,

$X_{L} = ω L = 2 π ν L = 2 π \times 50 \times 44 \times 1 0^{- 3} Ω$

Rms value of current,

$I_{r m s} = \frac{V _{r m s}}{X _{L}} = \frac{220}{2 π \times 50 \times 44 \times 1 0 ^{- 3}} = 15.92 A$

A $60 μ F$ capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. Sol. $C = 60 μ F = 60 \times 1 0^{- 6} F, V_{rms} = 110 V$ ,

$v = 60 Hz, ω = 2 π v$

Capacitive reactance,

$X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C} = \frac{1}{2 π \times 60 \times 60 \times 1 0 ^{- 6}} Ω X_{C} = 44.23Ω$

rms value of current,

$I_{r m s} = \frac{V _{r m s}}{X _{C}} = \frac{110}{44.23} = 2.49 A$

In Exercises 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. Sol. $P_{avg.} = V_{rms} I_{rms} cos ϕ$ For pure inductive circuit, $ϕ = \frac{π}{2}, P_{avg.} = 0$ For pure capacitive circuit, $ϕ = - \frac{π}{2}$

$P_{avg} = V_{rms} I_{rms} cos (- \frac{π}{2}) = 0$

Explanation : In each case power factor, $cos ϕ = 0$ , so $P_{avg}$ is zero.

A charged $30 μ F$ capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit ? Sol. $C = 30 μ F = 30 \times 1 0^{- 6} F$ , $L = 27 mH = 27 \times 1 0^{- 3} H$ Angular frequency,

$ω_{r} = \frac{1}{LC} = \frac{1}{27 \times 1 0 ^{- 3} \times 30 \times 1 0 ^{- 6}} = \frac{1}{9 \times 1 0 ^{- 4}} = 1.1 \times 1 0^{3} rad / s$

Hence, the angular frequency of free oscillations of the circuit is $1.11 \times 1 0^{3} rad / s$ .

A series LCR circuit with $R = 20Ω, L = 1.5 H$ and $C = 35 μ F$ is connected to a variable frequency 200 V ac 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. Here; $R = 20Ω, L = 1.5 H$ ,

$C = 35 μ F = 35 \times 1 0^{- 6} F, V_{rms} = 200 V$

At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Impedance of the circuit,

$Z = R^{2} + (X_{L} - X_{C})^{2}$

At resonance, $X_{L} = X_{C}$

$∴ Z = R = 20Ω$

Current in the circuit, $I_{rms} = \frac{V _{rms}}{Z} = \frac{200}{20} = 10 A$ Hence, the average power transferred to the circuit in one complete cycle :

$P_{avg} = V_{rms} I_{rms} cos ϕ$

$\Rightarrow P_{avg} = V_{ms} I_{ms} cos 0^{\circ} = 200 \times 10 \times 1 = 2000 W$ .

Figure shows a series LCR circuit connected to a variable frequency 230 V source. $L = 5.0 H, C = 80 μ F, R = 40Ω$ .

Series LCR circuit with resistor, inductor, and capacitor connected to a variable-frequency AC source.

(a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. Sol. Here;

$L C R V_{rms} = 5.0 H, = 80 μ F = 80 \times 1 0^{- 6} F, = 40Ω, = 230 V .$

(a) Resonant frequency,

$ω_{r} = \frac{1}{LC} = \frac{1}{5 \times 80 \times 1 0 ^{- 6}} = 50 rad / s$

(b) At resonant frequency, $X_{L} = X_{C}$ and

$\Rightarrow As such Z = R^{2} + (X_{L} - X_{C})^{2} Z = R = 40Ω$

Thus,

$I_{rms} = \frac{V _{rms}}{Z} = \frac{230}{40} = 5.75 A$

Amplitude of the current,

$I_{0} = I_{rms} 2 = 5.75 \times 2 = 8.1 A$

(c)

$(V_{rms})_{R} = I_{rms} \times R = 5.75 \times 40 = 230 V$

$(V_{rms})_{L} = I_{rms} X_{L} = I_{rms} ω_{r} L = 5.75 \times 50 \times 5 = 1437.5 V$

$(V_{rms}) = I_{rms} X_{C} = I_{rms} \times \frac{1}{ω _{r} C} = 5.75 (\frac{1}{50 \times ( 80 \times 1 0 ^{- 6} )}) Ω = 1437.5 V$

At resonance $X_{L} = X_{C}, (V_{rms})_{L} = (V_{rms})_{C}$

$(V_{rms})_{LC} = I_{rms} (X_{L} - X_{C}) = 0$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, important formulas, and simple explanations to help students learn concepts and practise questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 7

Clear Understanding of Phasor Diagrams: The solutions use phasor diagrams to explain the phase difference between voltage and current. This helps students understand why current leads or lags instead of simply memorising the relationships.
Step-by-Step LCR Circuit Solutions: Important questions on LCR circuits and resonance are explained step by step, including the derivation of the resonance frequency and related formulas.
Useful Comparison Tables: The solutions include tables to compare important quantities in AC circuits, such as resistance, inductive reactance, capacitive reactance, and their dependence on frequency.
Transformer Losses: The solutions explain the main causes of energy loss in transformers, including flux leakage, eddy currents, and copper losses, along with why these losses occur.
Easy Calculation of Circuit Values: The solutions explain how to calculate $V_{l} V_{c}$ and $V_{r}$ in an LCR circuit using phasor diagrams, vector addition, the Pythagorean theorem, and the impedance triangle.

NCERT Solutions Class 12 Physics Chapter 7 - Alternating Current

1.0Class 12 Physics Chapter 7 Alternating Current: Key Concepts

Class 12 Physics Chapter 7, Alternating Current, explains the behaviour of electrical circuits when the voltage and current change with time. The main concepts covered in this chapter include:

RMS and Peak Values: Understand the relationship between the peak value and RMS (effective) value of alternating current and voltage.
AC Through Resistor, Inductor and Capacitor: Learn how alternating current behaves in a resistor, inductor, and capacitor. In a resistor, current and voltage are in phase. In an inductor, current lags behind voltage by one-quarter cycle, while in a capacitor, current leads voltage by one-quarter cycle.
Series LCR Circuit: Study the combined behaviour of a resistor, inductor, and capacitor connected in series and learn how to calculate the total impedance of the circuit.
Resonance and Q-Factor: Understand resonance in an LCR circuit, where impedance is minimum and current is maximum. The chapter also explains the sharpness of resonance using the Q-factor.
Power in AC Circuits: Learn about power consumption in AC circuits, power factor, and wattless current.
LC Oscillations: Understand how energy is exchanged between an inductor and a capacitor in an LC circuit.
Transformers: Study how transformers use mutual induction to increase or decrease AC voltage and learn about the main causes of energy loss in transformers.

2.0NCERT Solutions Class 12 Physics Chapter 7 Alternating Current : Detailed Solutions

SOLVED EXAMPLES

A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source and (c) the rms current through the bulb Sol. (a) We are given $P = 100 W$ and $V = 220 V$ . The resistance of the bulb is

$R = \frac{V ^{2}}{P} = \frac{( 220 V ) ^{2}}{100 W} = 484Ω$

(b) The peak voltage of the source is

$V_{m} = 2 V = 100 W$

$I = \frac{P}{V} = \frac{100 W}{220 V} = 0.454 A$

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz . Sol. The inductive reactance,

$X_{L} = 2 π vL = 2 \times 3.14 \times 50 \times 25 \times 1 0^{- 3} Ω = 7.85Ω$

The rms current in the circuit is

$I = \frac{V}{X _{L}} = \frac{220 V}{7.85Ω} = 28 A$

A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Sol. When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitive reactance $(1/ ω C)$ and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.
A $15.0 μ F$ capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Sol. The capacitive reactance is

$X_{C} = \frac{1}{2 π ν C} = \frac{1}{2 \times 3.14 \times ( 50 Hz ) ( 15.0 \times 1 0 ^{- 6} F )} = 212Ω$

The rms current is $I = \frac{V}{X _{C}} = \frac{220 V}{212Ω} = 1.04 A$ The peak current is

$i_{m} = 2 I = (1.41) (1.04 A) = 1.47 A$

This current oscillates between +1.47A and -1.47 A , and is ahead of the voltage by $π /2$ . If the frequency is doubled, the capacitive reactance is halved and consequently, the current is double.

A light bulb and an open coil inductor are connected to an ac source through a key as shown in figure.

A resistor of $200Ω$ and a capacitor of $15.0 μ F$ are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Sol. Given : $R = 200Ω, C = 15.0 μ F = 15.0 \times 1 0^{- 6} F$

$V = 220 V, v = 50 Hz$

(a) In order to calculate the current, we need the impedance of the circuit. It is

$Z = R^{2} + X_{C}^{2} = R^{2} + (2 π ν C)^{2} = (200Ω)^{2} + (2 \times 3.14 \times 50 \times 15.0 \times 1 0^{- 6} F)^{2} = (200Ω)^{2} + (212.3Ω)^{2} = 291.67Ω$

Therefore, the current in the circuit is

$I = \frac{V}{Z} = \frac{220 V}{291.5Ω} = 0.755 A$

(b) Since the current is the same throughout the circuit, we have

$V_{R} = I R = (0.755 A) (200Ω) = 151 V V_{C} = I X_{C} = (0.755 A) (212.3Ω) = 160.3 V$

$V_{R + C} = V_{R}^{2} + V_{C}^{2} V_{R + C} = V_{R}^{2} + V_{C}^{2} = 220 V$

Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

(a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

Sol.

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which $R = 3Ω. L = 25.48 mH$ and $C = 796 μ F$ . Find (a) the impedance of the circuit ; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.

Sol.

(a) To find the impedance of the circuit, we first calculate $X_{L}$ and $X_{C}$ .

$X_{L} X_{C} = 2 π ν L = 2 \times 3.14 \times 50 \times 25.48 \times 1 0^{- 3} Ω = 8Ω = \frac{1}{2 π ν C} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 1 0 ^{- 6}} = 4Ω$

Therefore,

$Z = R^{2} + (X_{L} - X_{C})^{2} = 3^{2} + (8 - 4)^{2} = 5Ω$

(b) Phase difference,

$ϕ = tan^{- 1} \frac{X _{C} - X _{L}}{R} = tan^{- 1} (\frac{4 - 8}{3}) = - 53. 1^{\circ}$

Since $ϕ$ is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is $= I^{2} R$ Now,

$I = \frac{i _{m}}{2} = \frac{1}{2} (\frac{283}{5}) = 40 A$

Therefore, $P = (40 A)^{2} \times 3Ω = 4800 W$ (d) Power factor $= cos ϕ = cos (- 53. 1^{\circ}) = 0.6$

Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.

Sol.

(a) The frequency at which the resonance occurs is

$ω_{0} = \frac{1}{LC} = \frac{1}{25.48 \times 1 0 ^{- 3} \times 796 \times 1 0 ^{- 6}} = 222.1 rad / s v_{r} = \frac{ω _{0}}{2 π} = \frac{221.1}{2 \times 3.14} Hz = 35.4 Hz$

(b) The impedance Z at resonant condition is equal to the resistance :

$Z = R = 3Ω$

The rms current at resonance is,

$I = \frac{V}{Z} = \frac{V}{R} = (\frac{283}{2}) \times \frac{1}{3} = 66.7 A$

The power dissipated at resonance,

$P = I^{2} \times R = (66.7)^{2} \times 3 = 13.35 kW$

You can see that in the present case, power dissipated at resonance is more than the power dissipated in Question-8.

At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Sol. The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes - resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

EXERCISE QUESTIONS WITH SOLUTIONS

A $100Ω$ resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?

Sol. $R = 100Ω, V_{rms} = 220 V, f = 50 Hz$

(a) The rms value of current in the circuit,

$I_{rms} = \frac{V _{rms}}{R} = \frac{220}{100} = 2.20 A$

(b) The net power consumed over a full cycle,

$P_{avg} = V_{rms} I_{rms} cos ϕ = V_{rms} I_{rms} cos 0^{\circ} = 220 \times 2.2 \times 1 = 484 W$

(a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? Sol. (a) $V_{0} = 300 V$ rms voltage,

$V_{rms} = \frac{V _{0}}{2} = \frac{300}{2} = 212.16 V$

(b) The rms value of current, $I_{rms} = 10 A$ Now peak current,

$I_{0} = I_{rms} 2 = 102 = 14.1 A$

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. Sol. $L = 44 mH = 44 \times 1 0^{- 3} H$ ,

$V_{rms} = 220 V, v = 50 Hz$

Angular frequency, $ω = 2 π v$ Inductive reactance,

$X_{L} = ω L = 2 π ν L = 2 π \times 50 \times 44 \times 1 0^{- 3} Ω$

Rms value of current,

$I_{r m s} = \frac{V _{r m s}}{X _{L}} = \frac{220}{2 π \times 50 \times 44 \times 1 0 ^{- 3}} = 15.92 A$

A $60 μ F$ capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. Sol. $C = 60 μ F = 60 \times 1 0^{- 6} F, V_{rms} = 110 V$ ,

$v = 60 Hz, ω = 2 π v$

Capacitive reactance,

$X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C} = \frac{1}{2 π \times 60 \times 60 \times 1 0 ^{- 6}} Ω X_{C} = 44.23Ω$

rms value of current,

$I_{r m s} = \frac{V _{r m s}}{X _{C}} = \frac{110}{44.23} = 2.49 A$

In Exercises 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. Sol. $P_{avg.} = V_{rms} I_{rms} cos ϕ$ For pure inductive circuit, $ϕ = \frac{π}{2}, P_{avg.} = 0$ For pure capacitive circuit, $ϕ = - \frac{π}{2}$

$P_{avg} = V_{rms} I_{rms} cos (- \frac{π}{2}) = 0$

Explanation : In each case power factor, $cos ϕ = 0$ , so $P_{avg}$ is zero.

A charged $30 μ F$ capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit ? Sol. $C = 30 μ F = 30 \times 1 0^{- 6} F$ , $L = 27 mH = 27 \times 1 0^{- 3} H$ Angular frequency,

$ω_{r} = \frac{1}{LC} = \frac{1}{27 \times 1 0 ^{- 3} \times 30 \times 1 0 ^{- 6}} = \frac{1}{9 \times 1 0 ^{- 4}} = 1.1 \times 1 0^{3} rad / s$

Hence, the angular frequency of free oscillations of the circuit is $1.11 \times 1 0^{3} rad / s$ .

A series LCR circuit with $R = 20Ω, L = 1.5 H$ and $C = 35 μ F$ is connected to a variable frequency 200 V ac 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. Here; $R = 20Ω, L = 1.5 H$ ,

$C = 35 μ F = 35 \times 1 0^{- 6} F, V_{rms} = 200 V$

At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Impedance of the circuit,

$Z = R^{2} + (X_{L} - X_{C})^{2}$

At resonance, $X_{L} = X_{C}$

$∴ Z = R = 20Ω$

Current in the circuit, $I_{rms} = \frac{V _{rms}}{Z} = \frac{200}{20} = 10 A$ Hence, the average power transferred to the circuit in one complete cycle :

$P_{avg} = V_{rms} I_{rms} cos ϕ$

$\Rightarrow P_{avg} = V_{ms} I_{ms} cos 0^{\circ} = 200 \times 10 \times 1 = 2000 W$ .

Figure shows a series LCR circuit connected to a variable frequency 230 V source. $L = 5.0 H, C = 80 μ F, R = 40Ω$ .

$L C R V_{rms} = 5.0 H, = 80 μ F = 80 \times 1 0^{- 6} F, = 40Ω, = 230 V .$

(a) Resonant frequency,

$ω_{r} = \frac{1}{LC} = \frac{1}{5 \times 80 \times 1 0 ^{- 6}} = 50 rad / s$

(b) At resonant frequency, $X_{L} = X_{C}$ and

$\Rightarrow As such Z = R^{2} + (X_{L} - X_{C})^{2} Z = R = 40Ω$

Thus,

$I_{rms} = \frac{V _{rms}}{Z} = \frac{230}{40} = 5.75 A$

Amplitude of the current,

$I_{0} = I_{rms} 2 = 5.75 \times 2 = 8.1 A$

(c)

$(V_{rms})_{R} = I_{rms} \times R = 5.75 \times 40 = 230 V$

$(V_{rms})_{L} = I_{rms} X_{L} = I_{rms} ω_{r} L = 5.75 \times 50 \times 5 = 1437.5 V$

$(V_{rms}) = I_{rms} X_{C} = I_{rms} \times \frac{1}{ω _{r} C} = 5.75 (\frac{1}{50 \times ( 80 \times 1 0 ^{- 6} )}) Ω = 1437.5 V$

At resonance $X_{L} = X_{C}, (V_{rms})_{L} = (V_{rms})_{C}$

$(V_{rms})_{LC} = I_{rms} (X_{L} - X_{C}) = 0$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, important formulas, and simple explanations to help students learn concepts and practise questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 7

Clear Understanding of Phasor Diagrams: The solutions use phasor diagrams to explain the phase difference between voltage and current. This helps students understand why current leads or lags instead of simply memorising the relationships.
Step-by-Step LCR Circuit Solutions: Important questions on LCR circuits and resonance are explained step by step, including the derivation of the resonance frequency and related formulas.
Useful Comparison Tables: The solutions include tables to compare important quantities in AC circuits, such as resistance, inductive reactance, capacitive reactance, and their dependence on frequency.
Transformer Losses: The solutions explain the main causes of energy loss in transformers, including flux leakage, eddy currents, and copper losses, along with why these losses occur.
Easy Calculation of Circuit Values: The solutions explain how to calculate $V_{l} V_{c}$ and $V_{r}$ in an LCR circuit using phasor diagrams, vector addition, the Pythagorean theorem, and the impedance triangle.

Frequently Asked Questions

Which AC circuit concepts are explained through NCERT Solutions Class 12 Physics Chapter 7?

In an AC circuit, why does current lead or lag voltage according to NCERT Solutions Class 12 Physics Chapter 7?

Resonance occurs in a series LCR circuit—what does NCERT Solutions Class 12 Physics Chapter 7 explain?

Can NCERT Solutions Class 12 Physics Chapter 7 help solve questions based on power factor?

Why are transformers used in AC power transmission according to NCERT Solutions Class 12 Physics Chapter 7?

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NCERT Solutions Class 12 Physics Chapter 7 - Alternating Current

1.0Class 12 Physics Chapter 7 Alternating Current: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 7 Alternating Current : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 7

Table of Contents

Frequently Asked Questions

Which AC circuit concepts are explained through NCERT Solutions Class 12 Physics Chapter 7?

In an AC circuit, why does current lead or lag voltage according to NCERT Solutions Class 12 Physics Chapter 7?

Resonance occurs in a series LCR circuit—what does NCERT Solutions Class 12 Physics Chapter 7 explain?

Can NCERT Solutions Class 12 Physics Chapter 7 help solve questions based on power factor?

Why are transformers used in AC power transmission according to NCERT Solutions Class 12 Physics Chapter 7?

Are NCERT Solutions Class 12 Physics Chapter 7 useful for competitive exams?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 7 - Alternating Current

1.0Class 12 Physics Chapter 7 Alternating Current: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 7 Alternating Current : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 7

Table of Contents