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આ MCQ મોડ્યુલ આના પર આધારિત છે: Ac Inductor Capacitor
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Ac Inductor Capacitor
🎓 Class 12
Physics
CBSE
Theory
Ch 7 – Alternating Current
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Ac Inductor Capacitor
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Ac Inductor Capacitor
7.3 AC Voltage Applied to an Inductor
Consider an AC source connected to a pure inductor of self-inductance \(L\) with negligible resistance. Let the source voltage be \(v = v_m \sin\omega t\). The inductor opposes the change of current with a back-EMF \(\varepsilon = -L\,di/dt\).
Applying Kirchhoff's loop rule \(v - L\,di/dt = 0\) gives:
\(L\dfrac{di}{dt} = v_m \sin\omega t\)
Integrating both sides (and using the result that the integration constant is zero for a purely sinusoidal source):
\(i = -\dfrac{v_m}{\omega L}\cos\omega t = \dfrac{v_m}{\omega L}\sin\!\left(\omega t - \dfrac{\pi}{2}\right)\)
Comparing with \(i = i_m \sin(\omega t - \pi/2)\):
\(i_m = \dfrac{v_m}{\omega L} = \dfrac{v_m}{X_L}\)
Inductive Reactance: The quantity \(X_L = \omega L = 2\pi f L\) plays the role of "resistance" for an inductor in an AC circuit. SI unit: ohm (Ω). \(X_L\) increases with frequency - an inductor blocks high frequencies but allows DC (\(f=0 \Rightarrow X_L=0\)).
7.3.1 Phasor diagram for inductor
On the phasor diagram the V-phasor points along the +x-axis while the I-phasor is rotated 90° clockwise (lagging).
7.3.2 Average power in a pure inductor
Instantaneous power: \(p = vi = v_m i_m \sin\omega t \sin(\omega t - \pi/2) = -v_m i_m \sin\omega t \cos\omega t = -(v_m i_m/2)\sin 2\omega t\).
Average of \(\sin 2\omega t\) over a full cycle is zero, so:
\(\boxed{P_L = 0}\)
A pure inductor consumes no average power. Energy oscillates between the source and the magnetic field of the inductor - it is stored when current grows and returned when current falls.
7.4 AC Voltage Applied to a Capacitor
Now consider an AC source connected across a capacitor of capacitance \(C\) with no resistance.
Charge on the capacitor at any instant: \(q = Cv\). The current is the rate of flow of charge:
\(i = \dfrac{dq}{dt} = C\dfrac{dv}{dt} = C\dfrac{d}{dt}(v_m \sin\omega t) = \omega C v_m \cos\omega t\)
\(i = \dfrac{v_m}{1/\omega C}\sin\!\left(\omega t + \dfrac{\pi}{2}\right) = i_m \sin\!\left(\omega t + \dfrac{\pi}{2}\right)\)
Capacitive Reactance: \(X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}\). SI unit: ohm. \(X_C\) is large at low frequency (blocks DC) and small at high frequency (capacitor "shorts out" high-frequency signals).
7.4.1 Phasor diagram for capacitor
7.4.2 Average power in a pure capacitor
Following the same calculation as for L, average power \(P_C = 0\). Energy oscillates between source and electric field of the capacitor.
Mnemonic - "CIVIL": in a Capacitor, I leads V; V leads I in an inductor (L).
| Element | Opposition | Formula | Phase of i w.r.t. v | Avg. power |
|---|---|---|---|---|
| Resistor R | Resistance | R | in phase (0°) | I²R |
| Inductor L | Inductive reactance | XL = ωL | lags by 90° | 0 |
| Capacitor C | Capacitive reactance | XC = 1/(ωC) | leads by 90° | 0 |
Example 7.3 — Reactance of an inductor
A pure inductor of inductance 25 mH is connected to a 220 V (rms), 50 Hz source. Find (a) the inductive reactance and (b) the RMS current.
(a) \(X_L = 2\pi f L = 2\pi \times 50 \times 25 \times 10^{-3} = 7.85\) Ω.
(b) \(I = V/X_L = 220/7.85 = 28.0\) A. (Very high - real inductors include winding resistance.)
Example 7.4 — Reactance of a capacitor
A 15.0 μF capacitor is connected to a 220 V (rms), 50 Hz source. Find (a) XC and (b) RMS current.
(a) \(X_C = 1/(2\pi f C) = 1/(2\pi \times 50 \times 15\times 10^{-6}) = 212\) Ω.
(b) \(I = V/X_C = 220/212 = 1.04\) A.
Simulation: Reactance vs Frequency
Drag the frequency slider. Watch how \(X_L\) and \(X_C\) change with frequency for the chosen L and C.
| XL = 2π fL | 7.85 Ω |
| XC = 1/(2π fC) | 212.2 Ω |
| Behaviour at this frequency | capacitive (XC > XL) |
Activity 7.2 — Capacitor blocks DC, passes AC
- Connect a 1 μF capacitor in series with a torch bulb and a 6 V DC battery. Switch on.
- Replace the DC battery with a 6 V (rms) AC source at 50 Hz.
Predict: in which case does the bulb glow?
The bulb does not glow with DC because the capacitor blocks steady current (\(X_C = \infty\) at f = 0). With AC the bulb glows dimly - the capacitor offers a finite reactance \(X_C = 1/(2\pi\times 50\times 10^{-6}) \approx 3.2\) kΩ that allows alternating current to flow.
Competency-Based Questions L1L2L3L4L6
A radio engineer has two boxes: Box A contains a pure inductor L = 0.4 H; Box B contains a pure capacitor C = 100 μF. Each is to be tested separately with a 50 Hz, 220 V AC supply.
1. Inductive reactance of Box A is closest to: L1
(c) 126 Ω. \(X_L = 2\pi\times50\times0.4 = 125.7\) Ω.
2. State the phase difference between voltage and current for each box. L2
Box A (inductor): current lags voltage by 90°. Box B (capacitor): current leads voltage by 90°.
3. Calculate the RMS current drawn from Box B. L3
\(X_C = 1/(2\pi\times50\times100\times10^{-6}) = 31.8\) Ω. \(I = 220/31.8 = 6.92\) A.
4. Both boxes are now driven at 500 Hz instead of 50 Hz. Compare and contrast how the RMS current changes in each box. L4
For inductor: \(X_L\) becomes 10× larger ⇒ current becomes 10× smaller. For capacitor: \(X_C\) becomes 10× smaller ⇒ current becomes 10× larger. Inductors "throttle" high frequencies, capacitors pass them - basis of LC filters.
5. Propose a simple circuit using one of the boxes to "block" radio-frequency noise while passing 50 Hz mains. L6
Place Box A (the inductor) in series with the load. At 50 Hz, \(X_L\) is only 126 Ω (small drop). At 1 MHz, \(X_L = 2\pi\times10^6\times0.4 \approx 2.5\) MΩ - large enough to block almost all RF current. This is the principle of a line choke.
Assertion-Reason Questions
Options: (A) Both true, R explains A. (B) Both true, R does not explain A. (C) A true, R false. (D) A false, R true.
Assertion: A pure capacitor allows DC to flow continuously through it.
Reason: Capacitive reactance \(X_C = 1/(\omega C)\) is infinite at \(\omega = 0\).
(D). A is false (capacitor blocks DC after initial charging). R is true.
Assertion: A pure inductor connected to an AC source consumes no average power.
Reason: The current and voltage are 90° out of phase, making cos φ = 0.
(A). Both true and R is the correct explanation. Average power = VI cos φ = 0.
Assertion: Reactance and resistance both have the unit ohm but their physical meaning differs.
Reason: Resistance dissipates power; reactance only stores and returns energy.
(A). Both true; R explains A.
Frequently Asked Questions - Ac Inductor Capacitor
What is the main concept covered in Ac Inductor Capacitor?
In NCERT Class 12 Physics Chapter 7 (Alternating Current), "Ac Inductor Capacitor" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Ac Inductor Capacitor useful in real-life applications?
Real-life applications of "Ac Inductor Capacitor" from NCERT Class 12 Physics Chapter 7 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Ac Inductor Capacitor?
Key formulas in "Ac Inductor Capacitor" (NCERT Class 12 Physics Chapter 7 Alternating Current) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 7?
NCERT Class 12 Physics Chapter 7 (Alternating Current) is structured so each part builds on the previous one. "Ac Inductor Capacitor" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Ac Inductor Capacitor?
CBSE board questions from "Ac Inductor Capacitor" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Ac Inductor Capacitor" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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