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આ MCQ મોડ્યુલ આના પર આધારિત છે: Power Transformers
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Power Transformers
🎓 Class 12
Physics
CBSE
Theory
Ch 7 – Alternating Current
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Power Transformers
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Power Transformers
7.6 Power in AC Circuits — Power Factor
For a general LCR circuit driven by \(v = v_m \sin\omega t\) the current is \(i = i_m \sin(\omega t - \phi)\). The instantaneous power is \(p = vi\) and its time-average over one full cycle works out to:
\(P = V_{rms} I_{rms} \cos\phi\)
where \(V_{rms}\) and \(I_{rms}\) are the RMS source voltage and current, and \(\phi\) is the phase angle between them.
Power Factor: The quantity \(\cos\phi\) is the power factor of the circuit. \(\cos\phi = R/Z\). Three special cases:
- Pure R: \(\phi = 0 \Rightarrow \cos\phi = 1\) (maximum power transfer).
- Pure L or C: \(\phi = \pm 90° \Rightarrow \cos\phi = 0\) (no power consumed).
- LCR at resonance: \(\phi = 0 \Rightarrow \cos\phi = 1\).
7.6.1 Wattless Current
The current component \(I\sin\phi\) is perpendicular to V on the phasor diagram. It contributes zero average power — it is called the wattless current. The "in-phase" component \(I\cos\phi\) is the working current.
| Term | Formula | Unit | Significance |
|---|---|---|---|
| Apparent power | S = VI | VA (volt-ampere) | total source rating |
| Real (true) power | P = VI cos φ | W (watt) | actually dissipated |
| Reactive power | Q = VI sin φ | VAR (var) | oscillates between source and L/C |
| Power factor | cos φ = P/S = R/Z | dimensionless | 0 (worst) to 1 (best) |
Example 7.7 — Power in a lagging load
An LCR series circuit draws an RMS current of 5 A from a 230 V mains. The phase angle is 53° (current lagging). Find the (a) apparent power, (b) real power, (c) power factor.
(a) S = VI = 230 × 5 = 1150 VA.
(b) cos 53° = 0.6 ⇒ P = 1150 × 0.6 = 690 W.
(c) Power factor = 0.6 (lagging).
7.7 LC Oscillations
Charge a capacitor C (initial charge q₀) and short its terminals through an inductor L (no resistance). What happens? The capacitor discharges through L; current builds up; the magnetic energy stored in L feeds back to recharge C with opposite polarity; and the cycle continues. The system performs free LC oscillations at the natural angular frequency:
\(\omega_0 = \dfrac{1}{\sqrt{LC}}\)
Mathematically: applying Kirchhoff's loop rule \(q/C + L\,di/dt = 0\) and \(i = -dq/dt\):
\(\dfrac{d^2 q}{dt^2} + \dfrac{1}{LC}q = 0 \Rightarrow q(t) = q_0\cos\omega_0 t\)
Mechanical analogy: \(q \leftrightarrow x\); \(L \leftrightarrow m\); \(1/C \leftrightarrow k\). The total energy is constant:
\(U = \dfrac{q^2}{2C} + \dfrac{1}{2}Li^2 = \dfrac{q_0^2}{2C} = \text{constant}\)
In any real circuit a small resistance damps the oscillations — energy is gradually lost as heat.
7.8 Transformers
A transformer exploits Faraday's induction to step AC voltage up or down with high efficiency.
7.8.1 Construction
- Soft iron laminated core - guides magnetic flux, minimises eddy-current loss.
- Primary winding of Np turns - receives the input AC.
- Secondary winding of Ns turns - delivers the output AC.
7.8.2 Turns Ratio and Voltage Ratio
For an ideal transformer (lossless), the flux per turn is the same in both coils. By Faraday's law:
\(V_p = -N_p \dfrac{d\Phi}{dt},\quad V_s = -N_s\dfrac{d\Phi}{dt}\)
\(\boxed{\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = k\,\text{(turns ratio)}}\)
Energy conservation for an ideal transformer (input power = output power): \(V_p I_p = V_s I_s\), so
\(\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s} = \dfrac{1}{k}\)
If \(N_s > N_p\): voltage steps UP, current steps down → step-up transformer.
If \(N_s < N_p\): voltage steps DOWN, current steps up → step-down transformer.
| Type | Turns ratio | Voltage | Current | Typical use |
|---|---|---|---|---|
| Step-up | Ns > Np | increases | decreases | generating station → grid (e.g. 11 kV → 220 kV) |
| Step-down | Ns < Np | decreases | increases | distribution sub-station → home (11 kV → 230 V) |
| 1:1 (isolation) | Ns = Np | same | same | safety isolation (medical) |
7.8.3 Energy Losses in Real Transformers
- Flux leakage — not all primary flux links the secondary; reduced by interleaved windings.
- Copper loss (I²R) — wire resistance heats up; reduced by thick low-resistance copper.
- Iron / Eddy-current loss — induced currents in the core; reduced by using thin laminated sheets insulated from each other.
- Hysteresis loss — energy spent magnetising and demagnetising the iron each cycle; reduced by using soft iron / silicon steel.
- Humming — magnetostriction makes the core vibrate at twice mains frequency.
Example 7.8 — Transformer turns ratio
The primary of a transformer has 200 turns and the secondary 5000 turns. Input is 220 V at 5 A. Assuming the transformer is ideal, find the (a) output voltage, (b) output current, (c) output power.
(a) \(V_s = V_p \times N_s/N_p = 220 \times 5000/200 = 5500\) V (step-up).
(b) \(I_s = I_p \times N_p/N_s = 5 \times 200/5000 = 0.20\) A.
(c) Output power \(V_s I_s = 5500 \times 0.20 = 1100\) W = input power 220 × 5 (ideal).
Example 7.9 — Long-distance transmission
A power station produces 1 MW at 11 kV. It is to be transmitted 50 km on a line of total resistance 2 Ω. Compare the line losses if the voltage is transmitted as (a) 11 kV directly and (b) stepped up to 220 kV.
(a) At 11 kV: I = P/V = 10⁶/11000 = 90.9 A. Line loss = I²R = (90.9)²×2 = 16,530 W ≈ 1.65 % of 1 MW.
(b) At 220 kV: I = 10⁶/220000 = 4.55 A. Line loss = (4.55)²×2 = 41.4 W ≈ 0.004 %.
Stepping up reduces transmission loss by a factor of (220/11)² = 400!
Simulation: Transformer Ratio
Adjust the turns and the input voltage / current. See output voltage and current update according to the ideal transformer equations.
| Type | Step-up |
| Vs | 5500 V |
| Is | 0.20 A |
| P (in = out, ideal) | 1100 W |
Activity 7.4 — Improving the power factor with a capacitor
An industrial motor (inductive load) draws 10 A at 230 V with power factor 0.6 lagging. By connecting a capacitor of suitable value in parallel, the wattless current can be cancelled and the line current reduced.
Predict: will the real (useful) power change after adding the capacitor?
The real power stays the same (the motor still does the same work). But because cos φ → 1, the same real power is now delivered with a SMALLER line current. Reduced I² R losses in the cable; lower electricity bill - which is why factories install capacitor banks.
Competency-Based Questions L1L2L3L5L6
A village transformer steps 11 kV (primary RMS) down to 230 V (secondary RMS). It supplies an effective load of 50 kW. Assume the transformer is ideal.
1. The turns ratio Np:Ns equals: L1
(b) 48:1. 11000/230 = 47.8 ≈ 48.
2. Why must a transformer's input be alternating, not direct? L2
Mutual induction requires a CHANGING flux in the core. A steady DC produces a constant flux ⇒ no induced EMF in the secondary (after the brief switch-on transient). Worse - the heavy DC would saturate and burn out the primary.
3. Calculate the secondary RMS current. L3
Is = P/Vs = 50000/230 = 217 A.
4. Justify why the long-distance HT line uses 11 kV rather than 230 V. L5
Power = VI is fixed. Choosing high V keeps I small ⇒ I²R losses in the line are tiny. At 230 V the line would carry 217 A and lose huge amounts of heat in cable resistance; at 11 kV the line current is only 4.55 A.
5. Design a step-down transformer to convert 220 V household mains to 12 V (rms) to power an LED strip drawing 2 A. State the turns ratio and the primary current. L6
Np:Ns = 220:12 = 55:3 (or about 18.3:1). For ideal transformer, Pp = Ps = 12×2 = 24 W. Primary current Ip = 24/220 = 0.11 A. So pick e.g. 916 primary turns, 50 secondary turns on a small iron core.
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 transformer can step up DC voltage just like AC voltage.
Reason: The induced EMF in a coil depends on the rate of change of flux through it.
(D). A is false (DC ⇒ constant flux ⇒ no induced EMF). R is true and is in fact the reason A is false.
Assertion: Power factor of a purely inductive circuit is zero.
Reason: Phase angle is 90°.
(A). Both true and R explains A: cos 90° = 0.
Assertion: The core of a transformer is laminated.
Reason: Lamination reduces eddy-current losses by breaking the conducting cross-section.
(A). Both true; R explains A.
Frequently Asked Questions - Power Transformers
What is the main concept covered in Power Transformers?
In NCERT Class 12 Physics Chapter 7 (Alternating Current), "Power Transformers" 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 Power Transformers useful in real-life applications?
Real-life applications of "Power Transformers" 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 Power Transformers?
Key formulas in "Power Transformers" (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. "Power Transformers" 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 Power Transformers?
CBSE board questions from "Power Transformers" 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 "Power Transformers" 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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