Class 12 Physics Chapter 7 Important Questions (2026-27)
Important Questions for Class 12 Physics Chapter 7 – Alternating Current
Looking for Class 12 Physics Chapter 7 important questions? This page covers Alternating Current with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on LCR circuits, resonance, reactance and the transformer. Attempt them after finishing the NCERT chapter.
Very Short Answer Questions (1 Mark)
Power in AC Circuits
Question 1
The instantaneous current and voltage of an ac circuit are given by \(i = 10\sin 300t\ \text{A}\) and \(V = 200\sin 300t\ \text{V}\). What is the power dissipation in the circuit?
\(P = \dfrac{V_0 I_0}{2}\cos\phi = \dfrac{200 \times 10}{2} =\) 1000 W
Question 2
The instantaneous current and voltage of an ac circuit are given by \(i = 10\sin 314t\ \text{A}\) and \(v = 50\sin 314t\ \text{V}\). What is the power dissipation in the circuit?
\(P = \dfrac{V_0 I_0}{2} = \dfrac{50 \times 10}{2} =\) 250 W
Question 3
The instantaneous current and voltage of an ac circuit are given by \(i = 10\sin 314t\ \text{A}\) and \(v = 50\sin\left(314t + \dfrac{\pi}{2}\right)\ \text{V}\). What is the power dissipation?
Power dissipation = 0 (wattless current)
Question 4
Define the term 'wattless current'.
Question 5
For an ideal inductor connected across a sinusoidal ac voltage source, state which of the following quantities is zero: (i) Instantaneous power (ii) Average power over full cycle of the ac voltage source.
Capacitive Reactance and AC Circuits
Question 6
When an ac source is connected across an ideal inductor, show on a graph the nature of variation of the voltage and the current over one complete cycle.
Question 7
A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element when connected to a 210 V dc source?
Question 8
A heating element is marked 210 V, 630 W. Find the resistance of the element when connected to a 210 V dc source.
Question 9
Define capacitor reactance. Write its SI unit.
Question 10
A variable frequency ac source is connected to a capacitor. Will the displacement current change if the frequency of the ac source is decreased?
Question 11
Plot a graph showing variation of capacitive reactance with the change in the frequency of the ac source.
Question 12
Define 'quality factor' of resonance in a series LCR circuit. What is its SI unit?
Transformers
Question 13
Mention the two characteristic properties of the material suitable for making the core of a transformer.
Question 14
Why is the core of a transformer laminated?
Question 15
Why is the use of ac voltage preferred over dc voltage? Give two reasons.
Short Answer Questions (2–3 Marks)
Power in AC Circuits
Question 16
Prove that an ideal capacitor in an ac circuit does not dissipate power.
\(V = V_0\sin\omega t\), \(I = I_0\sin(\omega t + \pi/2)\)
Instantaneous power: \(P = VI = V_0 I_0 \sin\omega t \cos\omega t = \dfrac{V_0 I_0}{2}\sin 2\omega t\)
Average over a full cycle: \(\langle P \rangle = \dfrac{V_0 I_0}{2}\langle\sin 2\omega t\rangle = 0\). Thus the average power dissipated per cycle in a capacitor is zero.
Question 17
Prove that an ideal inductor does not dissipate power in an ac circuit.
\(V = V_0\sin\omega t\), \(I = I_0\sin(\omega t - \pi/2) = -I_0\cos\omega t\)
Instantaneous power: \(P = VI = -V_0 I_0 \sin\omega t\cos\omega t = -\dfrac{V_0 I_0}{2}\sin 2\omega t\)
Average over a cycle: \(\langle P \rangle = 0\).
Question 18
Derive an expression for the impedance of an ac circuit consisting of an inductor and a resistor.
From the phasor diagram: \(V = \sqrt{V_R^2 + V_L^2}\)
\(V = \sqrt{(IR)^2 + (IX_L)^2} = I\sqrt{R^2 + X_L^2}\)
Impedance: \(Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + \omega^2 L^2}\)
Phase angle: \(\tan\phi = \dfrac{X_L}{R}\).
Question 19
(i) For a given ac \(i = i_m\sin\omega t\), show that the average power dissipated in a resistor R over a complete cycle is \(\dfrac{1}{2}i_m^2 R\). (ii) A light bulb is rated at 100 W for a 220 V ac supply. Calculate the resistance of the bulb.
So \(\langle P \rangle = \dfrac{1}{2}i_m^2 R = I_{rms}^2 R\).
(ii) \(R = \dfrac{V_{rms}^2}{P} = \dfrac{(220)^2}{100} =\) 484 Ω
Question 20
(i) For a given ac \(i = i_m\sin\omega t\), show that the average power dissipated in a resistor R over a complete cycle is \(\dfrac{1}{2}i_m^2 R\). (ii) A light bulb is rated at 125 W for 250 V ac supply. Calculate the resistance of the bulb.
(ii) \(R = \dfrac{V^2}{P} = \dfrac{(250)^2}{125} =\) 500 Ω
Question 21
Two heating elements of resistances R₁ and R₂, when operated at a constant voltage V, consume powers P₁ and P₂ respectively. Deduce the expressions for the power of their combination when they are connected in (i) series and (ii) parallel across the same voltage supply.
(i) In series: \(R_s = R_1 + R_2 = V^2\left(\dfrac{1}{P_1} + \dfrac{1}{P_2}\right)\). Power \(P_s = \dfrac{V^2}{R_s} =\) \(\dfrac{P_1 P_2}{P_1 + P_2}\)
(ii) In parallel: \(\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{P_1 + P_2}{V^2}\). Power \(P_p = \dfrac{V^2}{R_p} =\) \(P_1 + P_2\)
RMS Values and Simple AC Calculations
Question 22
A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the rms current.
\(I_{rms} = \dfrac{V_{rms}}{X_C} = \dfrac{220}{212} \approx\) 1.04 A
Question 23
A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate (i) the resistance of the bulb and (ii) the rms current through the bulb.
(ii) \(I_{rms} = \dfrac{P}{V} = \dfrac{100}{220} =\) 0.45 A
Question 24
A light bulb is rated 200 W for 220 V ac supply of 50 Hz. Calculate (i) the resistance of the bulb and (ii) the rms current through the bulb.
(ii) \(I_{rms} = \dfrac{200}{220} \approx\) 0.91 A
Question 25
A light bulb is rated 150 W for 220 V ac supply of 60 Hz. Calculate (i) the resistance of the bulb and (ii) the rms current through the bulb.
(ii) \(I_{rms} = \dfrac{150}{220} \approx\) 0.68 A
Question 26
An alternating voltage given by \(V = 140\sin 314t\) is connected across a pure resistor of 50 Ω. Find (i) the frequency of the source, (ii) the rms current through the resistor.
\(V_{rms} = \dfrac{140}{\sqrt{2}} = 99\ \text{V}\). \(I_{rms} = \dfrac{99}{50} =\) 1.98 A
Question 27
An alternating voltage given by \(V = 280\sin 50\pi t\) is connected across a pure resistor of 40 Ω. Find (i) the frequency of the source, (ii) the rms current through the resistor.
(ii) \(V_{rms} = \dfrac{280}{\sqrt{2}} = 198\ \text{V}\). \(I_{rms} = \dfrac{198}{40} =\) 4.95 A
Question 28
An alternating voltage given by \(V = 70\sin 100\pi t\) is connected across a pure resistor of 25 Ω. Find (i) the frequency of the source, (ii) the rms current through the resistor.
(ii) \(V_{rms} = \dfrac{70}{\sqrt{2}} \approx 49.5\ \text{V}\). \(I_{rms} = \dfrac{49.5}{25} \approx\) 1.98 A
Capacitors and Inductors in AC Circuits
Question 29
An electric lamp having a coil of negligible inductance is connected in series with a capacitor and an ac source, and is glowing with a certain brightness. How does the brightness change on reducing (i) the capacitance and (ii) the frequency? Justify your answer.
Reducing C or reducing frequency \(\omega\) decreases the current, so the brightness of the lamp decreases in both cases.
Question 30
A lamp is connected in series with a capacitor. Predict your observation when this combination is connected across (i) an ac source and (ii) a dc battery. What change would you notice in each case if the capacitance is increased?
(ii) With ac: Current flows through the capacitor (\(X_C = 1/\omega C\)), and the lamp glows. Increasing C decreases \(X_C\), so brightness increases.
Question 31
A capacitor C, a variable resistor R and a bulb B are connected in series to the ac mains. The bulb glows with some brightness. How will the glow change if (i) a dielectric slab is introduced between the plates of the capacitor (keeping R the same), (ii) the resistance R is increased (keeping C the same)?
(ii) Increasing R increases the total impedance of the circuit, so the current decreases. Brightness decreases.
Question 32
An inductor L of inductance X_L is connected in series with a bulb B and an ac source. How would the brightness of the bulb change when (i) the number of turns in the inductor is reduced, (ii) an iron rod is inserted in the inductor, (iii) a capacitor of reactance X_C = X_L is inserted in series in the circuit? Justify your answer.
(ii) Iron core increases L and \(X_L\), decreasing current → brightness decreases. (iii) When \(X_C = X_L\), the total reactance cancels, leaving only R. The current becomes maximum, so brightness increases.
Question 33
(i) When an ac source is connected to an ideal inductor, show that the average power supplied by the source over a complete cycle is zero. (ii) A lamp is connected in series with an inductor and an ac source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor? Explain.
(ii) Brightness decreases, because inserting the iron rod increases the inductance L, which increases the inductive reactance \(X_L\), reducing the current.
Question 34
(a) When an ac source is connected to an ideal capacitor, show that the average power supplied by the source over a complete cycle is zero. (b) A lamp is connected in series with a capacitor. Predict your observations when the system is connected first across a dc and then an ac source. What happens in each case if the capacitance is reduced?
(b) With dc: the lamp glows momentarily during charging; once fully charged, the capacitor blocks current and the lamp does not glow. Reducing C has no appreciable effect on the steady state.
With ac: the lamp glows with the same brightness initially. If C is reduced, \(X_C\) increases and brightness decreases.
Question 35
The figure shows a series LCR circuit connected to a variable frequency 200 V source with L = 50 mH, C = 80 µF and R = 40 Ω. Determine (i) the source frequency which drives the circuit in resonance, (ii) the quality factor Q of the circuit.
\(f_0 = \dfrac{500}{2\pi} \approx\) 79.6 Hz
(ii) \(Q = \dfrac{\omega_0 L}{R} = \dfrac{500 \times 0.05}{40} =\) 0.625
Question 36
The figure shows a series LCR circuit connected to a variable frequency 250 V source with L = 40 mH, C = 100 µF and R = 50 Ω. Determine (i) the source frequency which drives the circuit in resonance, (ii) the quality factor Q.
\(f_0 = \dfrac{500}{2\pi} \approx\) 79.6 Hz
(ii) \(Q = \dfrac{\omega_0 L}{R} = \dfrac{500 \times 0.04}{50} =\) 0.4
Question 37
The figure shows a series LCR circuit with L = 5.0 H, C = 80 µF, R = 40 Ω connected to a variable frequency 240 V source. Calculate (i) the angular frequency of the source which drives the circuit at resonance, (ii) the current at the resonating frequency, (iii) the rms potential drop across the capacitor at resonance.
(ii) At resonance, \(Z = R\), so \(I_{rms} = \dfrac{240}{40} =\) 6 A
(iii) \(X_C = \dfrac{1}{\omega_0 C} = \dfrac{1}{50 \times 80 \times 10^{-6}} = 250\ \Omega\).
\(V_{rms,C} = I_{rms} X_C = 6 \times 250 =\) 1500 V
Question 38
A series LCR circuit with L = 4.0 H, C = 100 µF and R = 60 Ω is connected to a variable frequency 240 V source. Calculate (i) the angular frequency of the source at resonance, (ii) the current at resonance, (iii) the rms potential drop across the inductor at resonance.
(ii) \(I_{rms} = \dfrac{240}{60} =\) 4 A
(iii) \(X_L = \omega_0 L = 50 \times 4 = 200\ \Omega\). \(V_{rms,L} = I_{rms} X_L = 4 \times 200 =\) 800 V
Question 39
The figure shows a series LCR circuit with L = 10.0 H, C = 40 µF, R = 60 Ω connected to a variable frequency 240 V source. Calculate (i) the angular frequency of the source at resonance, (ii) the current at resonance, (iii) the rms potential drop across the inductor at resonance.
(ii) \(I_{rms} = \dfrac{240}{60} =\) 4 A
(iii) \(X_L = 50 \times 10 = 500\ \Omega\). \(V_{rms,L} = 4 \times 500 =\) 2000 V
Question 40
An inductor of unknown value, a capacitor of 100 µF and a resistor of 10 Ω are connected in series to a 200 V, 50 Hz ac source. It is found that the power factor of the circuit is unity. Calculate the inductance of the inductor and the current amplitude.
\(\omega L = \dfrac{1}{\omega C} \Rightarrow L = \dfrac{1}{\omega^2 C} = \dfrac{1}{(2\pi \times 50)^2 \times 100 \times 10^{-6}}\)
\(L = \dfrac{1}{314^2 \times 10^{-4}} = \dfrac{1}{9.87} \approx\) 0.101 H
At resonance, \(Z = R = 10\ \Omega\). \(I_{rms} = \dfrac{200}{10} = 20\ \text{A}\). Current amplitude \(I_0 = 20\sqrt{2} \approx\) 28.3 A
Question 41
A circuit containing an 80 mH inductor and a 250 µF capacitor in series is connected to a 240 V, 100 rad/s supply. The resistance of the circuit is negligible. (i) Obtain the rms value of current, (ii) What is the total average power consumed by the circuit?
\(X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 250 \times 10^{-6}} = 40\ \Omega\).
\(Z = X_C - X_L = 32\ \Omega\). \(I_{rms} = \dfrac{240}{32} =\) 7.5 A
(ii) Since R = 0, average power consumed = 0.
Question 42
A source of ac voltage \(V = V_0\sin\omega t\) is connected to a series combination of a resistor R and a capacitor C. Draw the phasor diagram and use it to obtain the expression for (i) impedance of the circuit and (ii) phase angle.
From the phasor diagram: \(V = \sqrt{V_R^2 + V_C^2} = I\sqrt{R^2 + X_C^2}\).
(i) Impedance: \(Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + \dfrac{1}{\omega^2 C^2}}\)
(ii) Phase angle: \(\tan\phi = \dfrac{X_C}{R} = \dfrac{1}{\omega RC}\) (current leads voltage).
Question 43
A 200 mH (pure) inductor and a 5 µF (pure) capacitor are connected, one by one, across a sinusoidal ac voltage source \(V = 70.7\sin(1000t)\). Obtain the expressions for the current in each case.
Inductor: \(X_L = \omega L = 1000 \times 0.2 = 200\ \Omega\). \(I_0 = \dfrac{70.7}{200} = 0.354\ \text{A}\).
Current lags voltage by \(\pi/2\): \(i_L = 0.354\sin(1000t - \pi/2) = \) \(-0.354\cos(1000t)\ \text{A}\)
Capacitor: \(X_C = \dfrac{1}{\omega C} = \dfrac{1}{1000 \times 5 \times 10^{-6}} = 200\ \Omega\). \(I_0 = 0.354\ \text{A}\).
Current leads voltage by \(\pi/2\): \(i_C = 0.354\sin(1000t + \pi/2) = \) \(0.354\cos(1000t)\ \text{A}\)
Transformers
Question 44
State the principle of working of a transformer. Can a transformer be used to step up or step down a dc voltage? Justify your answer.
No, a transformer cannot be used with dc. A dc voltage cannot produce a change in magnetic flux, which is required to induce an emf in the secondary.
Question 45
Mention various energy losses in a transformer.
Question 46
State the underlying principle of a transformer. How is the large scale transmission of electric energy over long distances done with the use of transformers?
Long-distance transmission: The generator output is stepped up (to very high voltage) so that the current in the transmission lines is greatly reduced. This reduces the I²R power loss in the wires. At the consumer end, the voltage is stepped down at substations to a safe, usable level.
Question 47
A step-up transformer converts a low voltage into a high voltage. Does it violate the principle of conservation of energy? Explain.
Question 48
(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step-up transformer. (ii) Write any two sources of energy loss in a transformer.
(ii) Two sources of energy loss: (a) Copper loss (I²R in windings), (b) Eddy current loss (in the core) — or hysteresis loss.
Question 49
(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown. Which one leads in phase: current or voltage? (ii) Without making any other change, find the value of the additional capacitor C' to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.
If \(X_C > X_L\), current leads voltage. Phase angle: \(\tan\phi = \dfrac{X_C - X_L}{R}\).
(ii) To make power factor unity, need \(X_C' = X_L\), so new capacitance \(C'_{total} = \dfrac{1}{\omega X_L}\). The additional capacitor \(C' = C'_{total} - C\), connected in parallel.
Question 50
(a) Determine the value of phase difference between the current and the voltage in the given series LCR circuit. (b) Calculate the value of the additional capacitor which may be joined suitably to the capacitor C that would make the power factor of the circuit unity.
Question 51
(a) Draw a schematic arrangement for winding of primary and secondary coil in a transformer when the two coils are wound on top of each other. (b) State the underlying principle of a transformer and obtain the expression for the ratio of secondary to primary voltage in terms of the (i) number of secondary and primary windings and (ii) primary and secondary currents. (c) Write the main assumption involved in deriving the above relations. (d) Write any two reasons due to which energy losses may occur in actual transformers.
(b) Principle: mutual induction. If flux linked per turn is the same for both coils:
\(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}\) (the last equality from \(V_p I_p = V_s I_s\) for an ideal transformer).
(c) Main assumptions: (i) Primary resistance is small, (ii) Same flux links both coils (no flux leakage).
(d) Two energy losses: copper loss and eddy current loss (or hysteresis/flux leakage).
Question 52
(i) Draw a labelled diagram of a step-down transformer. State the principle of its working. (ii) Express the turn ratio in terms of voltages. (iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer. (iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V – 550 W refrigerator?
(ii) Turn ratio: \(\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}\).
(iii) For ideal transformer: \(\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}\).
(iv) Output power = 550 W. Input power = 550 W (ideal). Primary current: \(I_p = \dfrac{550}{220} =\) 2.5 A
Question 53
(i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device. (ii) The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is 100. The input voltage and power are respectively 220 V and 1100 W. Calculate (a) number of turns in secondary, (b) current in primary, (c) voltage across secondary, (d) current in secondary, (e) power in secondary.
(ii) Transformation ratio = \(\dfrac{N_s}{N_p} = 100\).
(a) \(N_s = 100 \times N_p = 100 \times 100 =\) 10,000 turns
(b) \(I_p = \dfrac{P}{V_p} = \dfrac{1100}{220} =\) 5 A
(c) \(V_s = 100 \times V_p = 100 \times 220 =\) 22,000 V
(d) \(I_s = \dfrac{P}{V_s} = \dfrac{1100}{22000} =\) 0.05 A
(e) Power in secondary = \(V_s I_s\) = 1100 W (ideal transformer)
Long Answer Questions (5 Marks)
AC Through a Capacitor
Question 54
An ac source generating a voltage \(v = v_m\sin\omega t\) is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of v and i versus t to show that the current is \(\pi/2\) ahead of the voltage. A resistor of 200 Ω and a capacitor of 15.0 µF are connected in series to a 220 V, 50 Hz ac source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.
\(i = \dfrac{dq}{dt} = \omega C v_m\cos\omega t = i_m\sin(\omega t + \pi/2)\)
So current leads the voltage by \(\pi/2\) — the graph shows sinusoidal v and i with i ahead of v by a quarter cycle.
Numerical: \(X_C = \dfrac{1}{2\pi \times 50 \times 15 \times 10^{-6}} = 212\ \Omega\).
\(Z = \sqrt{R^2 + X_C^2} = \sqrt{200^2 + 212^2} \approx 291\ \Omega\).
\(I_{rms} = \dfrac{220}{291} =\) 0.756 A
\(V_{rms,R} = I_{rms}R = 0.756 \times 200 = 151.2\ \text{V}\)
\(V_{rms,C} = I_{rms}X_C = 0.756 \times 212 = 160.3\ \text{V}\)
Sum \(= 311.5\ \text{V}\), which is greater than 220 V. These voltages are \(\pi/2\) out of phase, so they must be added as phasors: \(V = \sqrt{V_R^2 + V_C^2} = 220\ \text{V}\) — no paradox.
Average Power in LCR Circuit
Question 55
A voltage \(V = V_0\sin\omega t\) is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though current flows, (ii) maximum power dissipated?
Instantaneous power: \(P = VI = V_0 I_0 \sin\omega t \sin(\omega t - \phi)\)
Using trig identity and averaging over a cycle: \(\langle\sin^2\rangle = 1/2\), \(\langle\sin\cos\rangle = 0\):
Average power: \(P = \dfrac{V_0 I_0}{2}\cos\phi = V_{rms} I_{rms}\cos\phi\)
(i) No power dissipated when \(\phi = \pi/2\) (\(X_L = X_C\) impossible; rather when circuit is purely inductive or purely capacitive, \(\cos\phi = 0\)) — wattless current.
(ii) Maximum power when \(\phi = 0\) (resonance, \(X_L = X_C\)), \(\cos\phi = 1\): \(P_{max} = I_{rms}^2 R\).
Question 56
Derive the expression for the average power dissipated in a series LCR circuit for an ac source of voltage \(v = v_m\sin\omega t\), carrying a current \(i = i_m\sin(\omega t + \phi)\). Hence define 'wattless current'. State under what condition it can be realized in a circuit.
Wattless current: The component of the current in an ac circuit that consumes no power. It occurs when the current and voltage differ in phase by \(\pi/2\) (in a pure inductor or pure capacitor, \(\cos\phi = 0\)).
Question 57
A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
\(V = \sqrt{V_R^2 + (V_L - V_C)^2} = I\sqrt{R^2 + (X_L - X_C)^2}\)
\(Z = \sqrt{R^2 + (X_L - X_C)^2}\), \(\tan\phi = \dfrac{X_L - X_C}{R}\)
Graph I vs f: I is small at both low and high frequencies and reaches a sharp maximum at the resonant frequency \(f_0 = \dfrac{1}{2\pi\sqrt{LC}}\). At \(f_0\), \(X_L = X_C\), \(Z = R\) (minimum), and I is maximum. The sharpness of the peak depends on R — smaller R gives sharper peak.
Question 58
In a series LCR circuit connected to an ac source of variable frequency and voltage \(v = V_m\sin\omega t\), draw a plot showing the variation of current I with angular frequency \(\omega\) for two different values of resistance R₁ and R₂ (R₁ > R₂). Write the condition under which resonance occurs. For which value of resistance is a sharper resonance produced? Define Q-factor and give its significance.
Condition for resonance: \(X_L = X_C\), i.e., \(\omega_0 = \dfrac{1}{\sqrt{LC}}\), and \(Z = R\).
Sharper resonance: For \(R_2\) (smaller R), the peak is sharper and higher.
Q-factor: \(Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}\).
Significance: Higher Q means a narrower and sharper resonance — the circuit is more selective (better at picking out one frequency, as in radio tuning).
Question 59
A series LCR circuit is connected to an ac source having voltage \(v = v_m\sin\omega t\). Derive the expression for the instantaneous current I and its phase relationship to the applied voltage. Obtain the condition for resonance. Define 'power factor'. State the conditions under which it is (i) maximum and (ii) minimum.
Resonance: When \(X_L = X_C\), i.e., \(\omega_0 = \dfrac{1}{\sqrt{LC}}\), then \(\phi = 0\), \(Z = R\), \(I_0\) maximum.
Power factor: \(\cos\phi\).
(i) Maximum: \(\cos\phi = 1\), when the circuit is purely resistive (\(\phi = 0\)).
(ii) Minimum: \(\cos\phi = 0\), when the circuit is purely inductive or purely capacitive (\(\phi = \pm\pi/2\)).
Question 60
(a) Derive an expression for the average power consumed in a series LCR circuit connected to an ac source where the phase difference between the voltage and current is zero. (b) Define the quality factor in an ac circuit. Why should the quality factor have a high value in receiving circuits? Name the factors on which it depends.
(b) Q-factor: \(Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}\).
Why high Q: A high Q means sharper resonance — the circuit is more selective, favoring one particular frequency (the received radio signal) and rejecting others.
Depends on: f₀ (resonant frequency), L, R and C.
Question 61
(a) Derive the relationship between the peak and rms value of current in an ac circuit. (b) Describe briefly, with the help of a labelled diagram, the working of a step-up transformer. A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.
\(I_{rms}^2 = \dfrac{1}{T}\int_0^T I_m^2\sin^2\omega t\,dt = \dfrac{I_m^2}{2}\). So \(I_{rms} = \dfrac{I_m}{\sqrt{2}} \approx 0.707 I_m\)
(b) Same principle (mutual induction) and working as in previous transformer questions. \(N_s > N_p\), so \(V_s > V_p\).
No violation: In an ideal transformer, \(V_p I_p = V_s I_s\) — what is gained in voltage is lost in current. Energy is conserved.
AC Generator
Question 62
Draw a sketch showing the basic elements of an ac generator. State its principle and explain briefly its working.
Elements: (i) Field magnet, (ii) Armature (coil), (iii) Slip rings, (iv) Brushes.
Working: Rotating the coil changes the flux through it. Every half rotation, the direction of induced emf reverses (since the coil sides swap their positions relative to the field). The slip rings allow the induced emf to be drawn out into the external circuit while the coil rotates.
Emf: \(\varepsilon = NAB\omega\sin\omega t\) where \(\omega\) is the angular velocity, so \(\varepsilon_0 = NAB\omega\).
Question 63
Explain briefly, with the help of a labelled diagram, the basic principle of the working of an ac generator. In an ac generator, a coil of N turns and area A is rotated at v revolutions per second in a uniform magnetic field B. Write the expression for the emf produced. A 100-turn coil of area 0.1 m² rotates at half a revolution per second. It is placed in a magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil.
Calculation: \(N = 100\), \(A = 0.1\ \text{m}^2\), \(B = 0.01\ \text{T}\), \(\nu = 0.5\ \text{rev/s}\), \(\omega = 2\pi\nu = \pi\ \text{rad/s}\).
Maximum emf: \(\varepsilon_0 = NAB\omega = 100 \times 0.1 \times 0.01 \times \pi =\) 0.314 V
Question 64
Describe briefly, with the help of a labelled diagram, the basic elements of an ac generator. State its underlying principle. Show diagrammatically how an alternating emf is generated by a loop of wire rotating in a magnetic field. Write the expression for the instantaneous value of the emf induced in the rotating loop.
\(\varepsilon = NAB\omega\sin\omega t\).
Graph: The emf vs time graph is a sinusoidal curve with maximum value \(NAB\omega\), passing through zero twice per cycle and flipping sign each half cycle.
Question 65
State the working of ac generator with the help of a labelled diagram. The coil of an ac generator having N turns, each of area A, is rotated with a constant angular velocity \(\omega\). Deduce the expression for the alternating emf generated in the coil. What is the source of energy generation in this device?
\(\phi = NAB\cos\omega t\), \(\varepsilon = -\dfrac{d\phi}{dt} = NAB\omega\sin\omega t\).
Source of energy: The mechanical energy of whatever drives the rotation of the coil (e.g., water turbine, steam turbine) is converted into electrical energy.
Question 66
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of (i) magnetic flux and (ii) alternating emf versus time generated by a loop of wire rotating in a magnetic field. (b) Why is a choke coil needed in the use of fluorescent tubes with ac mains?
(i) Magnetic flux: \(\phi = NAB\cos\omega t\) — a cosine curve, maximum when the coil is perpendicular to B, zero when parallel.
(ii) Emf: \(\varepsilon = NAB\omega\sin\omega t\) — a sine curve, maximum when flux is zero, zero when flux is maximum.
(b) A choke coil is used to control the current in an ac circuit (fluorescent tube) without dissipating much power. Unlike a resistor, a choke coil has negligible resistance and hence produces almost no heat loss — it controls current through its large inductive reactance instead.
Question 67
(a) Draw a labelled diagram of an ac generator and state its working principle. (b) How is magnetic flux linked with the armature coil changed in a generator? (c) Derive the expression for the maximum value of the induced emf and state the rule that gives the direction of the induced emf. (d) Show the variation of the emf generated versus time as the armature is rotated with respect to the direction of the magnetic field.
(b) As the coil rotates, the angle between the normal to the coil and the magnetic field changes, so the flux \(\phi = NAB\cos\omega t\) changes with time.
(c) From Faraday's law: \(\varepsilon = -\dfrac{d\phi}{dt} = NAB\omega\sin\omega t\). Maximum: \(\varepsilon_0 = NAB\omega\). Direction by Fleming's right-hand rule.
(d) Plot \(\varepsilon\) vs t: a sinusoidal wave with maxima at \(\pm NAB\omega\) and zeros at the times the coil plane is perpendicular to B.
Question 68
Discuss how Faraday's law of electromagnetic induction is applied in an ac generator for converting mechanical energy into electrical energy. Obtain an expression for the instantaneous value of the induced emf in an ac generator. Draw graphs to show the phase relationship between (i) magnetic flux linked with the coil and (ii) induced emf in the coil.
With \(\phi = NAB\cos\omega t\): \(\varepsilon = NAB\omega\sin\omega t\).
Graphs: Flux is a cosine curve starting at max. Emf is a sine curve (zero when flux max, max when flux zero) — so they are \(90^\circ\) out of phase.
Key Formulas to Remember
| Concept | Formula |
|---|---|
| RMS current | \(I_{rms} = \dfrac{I_0}{\sqrt{2}}\) |
| Inductive reactance | \(X_L = \omega L\) |
| Capacitive reactance | \(X_C = \dfrac{1}{\omega C}\) |
| Impedance of LCR | \(Z = \sqrt{R^2 + (X_L - X_C)^2}\) |
| Phase angle | \(\tan\phi = \dfrac{X_L - X_C}{R}\) |
| Power factor | \(\cos\phi = \dfrac{R}{Z}\) |
| Average power | \(P = V_{rms} I_{rms}\cos\phi\) |
| Resonant frequency | \(f_0 = \dfrac{1}{2\pi\sqrt{LC}}\), \(\omega_0 = \dfrac{1}{\sqrt{LC}}\) |
| Quality factor | \(Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}\) |
| Transformer (ideal) | \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}\) |
| EMF of ac generator | \(\varepsilon = NAB\omega\sin\omega t\), \(\varepsilon_0 = NAB\omega\) |
FAQs
Are these questions enough for CBSE Class 12 Physics Chapter 7?
Yes. These questions cover every major topic — ac through resistors, inductors and capacitors, LCR circuits, resonance, power factor, transformers and ac generators — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.
Which topics are most important for board exams?
Average power in ac circuits (including wattless current), phasor treatment of LCR circuits (impedance and phase angle), resonance and Q-factor, transformer (principle, working, energy losses, relation between V, N, I), and ac generator (principle, expression for emf) are consistently high-weightage topics.
How should I prepare for numericals in this chapter?
Practise computations of X_L, X_C, Z, phase angle, rms currents, resonance frequency, Q-factor, transformer voltage-current ratios, and average power. Focus on distinguishing when to use rms vs peak values, and on proper phasor treatment where voltages are out of phase.