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Electromagnetic Induction and Alternating Currents - Transformers and AC Generators

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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AC Generator Principle: An AC generator converts mechanical energy into electrical energy. It works on the principle of electromagnetic induction, where a coil rotated in a uniform magnetic field B⃗\vec{B} experiences a change in magnetic flux ϕ\phi, thereby inducing an alternating electromotive force (emf).

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Transformer Principle: A transformer is a device used to increase or decrease the amplitude of alternating voltage. It operates on the principle of mutual induction between two coils (primary and secondary) wound over a common laminated iron core.

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Step-up Transformer: A transformer that increases the output voltage (Vs>VpV_s > V_p) by having more turns in the secondary coil than the primary coil (Ns>NpN_s > N_p).

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Step-down Transformer: A transformer that decreases the output voltage (Vs<VpV_s < V_p) by having fewer turns in the secondary coil than the primary coil (Ns<NpN_s < N_p).

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Energy Losses in Transformers: Energy is lost during the process due to Copper loss (I2RI^2R heating), Eddy current loss (minimized by using a laminated core), Hysteresis loss (minimized by using soft iron), and Flux leakage.

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Ideal Transformer: In an ideal transformer, there is no loss of energy, meaning input power equals output power (VpIp=VsIsV_p I_p = V_s I_s).

📐Formulae

e=NBAωsin⁡(ωt)e = N B A \omega \sin(\omega t) (Instantaneous emf in AC generator)

e0=NBAωe_0 = N B A \omega (Peak emf in AC generator)

\frac{V_s}{V_p} = rac{N_s}{N_p} = k (Transformer turns ratio)

IpIs=VsVp=NsNp\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p} (Current-Voltage relation for ideal transformer)

η=PoutPin×100%=VsIsVpIp×100%\eta = \frac{P_{out}}{P_{in}} \times 100\% = \frac{V_s I_s}{V_p I_p} \times 100\% (Efficiency of a transformer)

💡Examples

Problem 1:

A transformer has 200200 turns in the primary and 40004000 turns in the secondary. If the primary is connected to an AC source of 220 V220\text{ V}, calculate the secondary voltage and the turns ratio kk.

Solution:

Given: Np=200N_p = 200, Ns=4000N_s = 4000, Vp=220 VV_p = 220\text{ V}. Turns ratio k=NsNp=4000200=20k = \frac{N_s}{N_p} = \frac{4000}{200} = 20. Using the formula VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}, we have: Vs=Vp×NsNp=220×20=4400 VV_s = V_p \times \frac{N_s}{N_p} = 220 \times 20 = 4400\text{ V}.

Explanation:

Since Ns>NpN_s > N_p, this is a step-up transformer, which results in a secondary voltage higher than the primary voltage.

Problem 2:

A rectangular coil of 100100 turns and area 0.2 m20.2\text{ m}^2 rotates at a frequency of 50 Hz50\text{ Hz} in a uniform magnetic field of 0.05 T0.05\text{ T}. Calculate the maximum emf induced in the coil.

Solution:

Given: N=100N = 100, A=0.2 m2A = 0.2\text{ m}^2, B=0.05 TB = 0.05\text{ T}, f=50 Hzf = 50\text{ Hz}. Angular velocity ω=2πf=2×π×50=100π rad/s\omega = 2 \pi f = 2 \times \pi \times 50 = 100\pi\text{ rad/s}. Maximum emf e0=NBAωe_0 = NBA\omega: e0=100×0.05×0.2×100π=100π≈314.16 Ve_0 = 100 \times 0.05 \times 0.2 \times 100\pi = 100\pi \approx 314.16\text{ V}.

Explanation:

The peak emf occurs when the plane of the coil is parallel to the magnetic field, making sin⁡(ωt)=1\sin(\omega t) = 1.

Transformers and AC Generators - Revision Notes & Key Formulas | ICSE Class 12 Physics