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Alternating Current - Transformers

Grade 12CBSEPhysics

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

🔑Concepts

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A transformer is an electrical device based on the principle of mutual induction used to increase or decrease the alternating voltage.

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It consists of two coils: the primary coil (input) and the secondary coil (output), wound on a laminated soft iron core.

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The principle of operation relies on Faraday's Law of Induction: a changing current in the primary coil creates a changing magnetic flux, which induces an emf in the secondary coil.

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In a Step-up Transformer, the number of turns in the secondary coil NsN_s is greater than the primary coil NpN_p (Ns>NpN_s > N_p), resulting in Vs>VpV_s > V_p and Is<IpI_s < I_p.

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In a Step-down Transformer, the number of turns in the secondary coil NsN_s is less than the primary coil NpN_p (Ns<NpN_s < N_p), resulting in Vs<VpV_s < V_p and Is>IpI_s > I_p.

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Energy losses in a real transformer include Copper loss (I2RI^2R heating), Eddy current loss (minimized by lamination), Hysteresis loss (minimized by using soft iron), and Flux leakage.

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For an ideal transformer, there is no power loss, meaning input power equals output power: VpIp=VsIsV_p I_p = V_s I_s.

📐Formulae

VsVp=NsNp=K\frac{V_s}{V_p} = \frac{N_s}{N_p} = K

For an ideal transformer: IpIs=VsVp=NsNp\text{For an ideal transformer: } \frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}

Efficiency (η)=Output PowerInput Power×100%=VsIsVpIp×100%\text{Efficiency } (\eta) = \frac{\text{Output Power}}{\text{Input Power}} \times 100\% = \frac{V_s I_s}{V_p I_p} \times 100\%

Induced emf in primary: εp=−Npdϕdt\text{Induced emf in primary: } \varepsilon_p = -N_p \frac{d\phi}{dt}

Induced emf in secondary: εs=−Nsdϕdt\text{Induced emf in secondary: } \varepsilon_s = -N_s \frac{d\phi}{dt}

💡Examples

Problem 1:

A step-down transformer has a primary coil with 20002000 turns and a secondary coil with 100100 turns. If the primary voltage is 2200 V2200\text{ V} and the output current is 10 A10\text{ A}, calculate the secondary voltage and the primary current, assuming the transformer is 100%100\% efficient.

Solution:

Given: Np=2000N_p = 2000, Ns=100N_s = 100, Vp=2200 VV_p = 2200\text{ V}, Is=10 AI_s = 10\text{ A}.

  1. To find VsV_s: Using the turn ratio formula VsVp=NsNp  ⟹  Vs=Vp(NsNp)=2200×(1002000)=110 V\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = V_p \left(\frac{N_s}{N_p}\right) = 2200 \times \left(\frac{100}{2000}\right) = 110\text{ V}.
  2. To find IpI_p: For an ideal transformer VpIp=VsIs  ⟹  Ip=VsIsVp=110×102200=0.5 AV_p I_p = V_s I_s \implies I_p = \frac{V_s I_s}{V_p} = \frac{110 \times 10}{2200} = 0.5\text{ A}.

Explanation:

Since it is a step-down transformer, the voltage decreases from 2200 V2200\text{ V} to 110 V110\text{ V} while the current increases from 0.5 A0.5\text{ A} in the primary to 10 A10\text{ A} in the secondary to conserve power.

Problem 2:

A transformer has an efficiency of 90%90\%. It is used to deliver 4.5 kW4.5\text{ kW} of power to a device at 220 V220\text{ V}. If the primary voltage is 1100 V1100\text{ V}, find the primary current.

Solution:

Given: η=0.90\eta = 0.90, Pout=4.5 kW=4500 WP_{out} = 4.5\text{ kW} = 4500\text{ W}, Vp=1100 VV_p = 1100\text{ V}. Efficiency is defined as η=PoutPin  ⟹  0.90=4500Pin  ⟹  Pin=45000.90=5000 W\eta = \frac{P_{out}}{P_{in}} \implies 0.90 = \frac{4500}{P_{in}} \implies P_{in} = \frac{4500}{0.90} = 5000\text{ W}. Now, Pin=VpIp  ⟹  5000=1100×Ip  ⟹  Ip=50001100≈4.55 AP_{in} = V_p I_p \implies 5000 = 1100 \times I_p \implies I_p = \frac{5000}{1100} \approx 4.55\text{ A}.

Explanation:

In a real transformer, the input power must be higher than the output power to account for energy losses. The primary current is calculated based on this total input power requirement.