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Electricity and Magnetism - Transformers

Grade 11IGCSEPhysics

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

πŸ”‘Concepts

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A transformer is a device used to increase or decrease the VoutV_{out} (output voltage) of an alternating current (ACAC). It consists of a primary coil, a secondary coil, and a soft iron core.

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The core is made of soft iron because it is easily magnetized and demagnetized, which helps to concentrate the magnetic field lines and improve flux linkage.

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Transformers work on the principle of mutual induction: an ACAC in the primary coil creates a changing magnetic field, which induces a varying EMFEMF (voltage) in the secondary coil.

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A Step-up transformer increases voltage (Vs>VpV_s > V_p) and has more turns on the secondary coil (Ns>NpN_s > N_p).

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A Step-down transformer decreases voltage (Vs<VpV_s < V_p) and has fewer turns on the secondary coil (Ns<NpN_s < N_p).

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In an ideal transformer, there is 100%100\% efficiency, meaning the input power equals the output power: Pin=PoutP_{in} = P_{out}.

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For long-distance power transmission, step-up transformers are used to increase voltage. Since P=VIP = VI, increasing voltage decreases the current II. This reduces power loss in the cables, which is calculated as Ploss=I2RP_{loss} = I^2 R.

πŸ“Formulae

VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}

V_p I_p = V_s I_s \text{ (for 100% efficiency)}

P=VIP = VI

Ploss=I2RP_{loss} = I^2 R

Efficiency=PoutputPinputΓ—100%=VsIsVpIpΓ—100%\text{Efficiency} = \frac{P_{output}}{P_{input}} \times 100\% = \frac{V_s I_s}{V_p I_p} \times 100\%

πŸ’‘Examples

Problem 1:

A transformer has 200200 turns on the primary coil and 40004000 turns on the secondary coil. If the input voltage is 230Β V230\text{ V}, calculate the output voltage.

Solution:

VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s} 230Vs=2004000\frac{230}{V_s} = \frac{200}{4000} Vs=230Γ—4000200V_s = \frac{230 \times 4000}{200} Vs=4600Β VV_s = 4600\text{ V}

Explanation:

By using the transformer turns ratio formula, we substitute the known values of NpN_p, NsN_s, and VpV_p to find that the secondary voltage VsV_s is 4600Β V4600\text{ V}. This is a step-up transformer.

Problem 2:

An ideal transformer steps down a 2400Β V2400\text{ V} supply to 120Β V120\text{ V}. If the current in the secondary circuit is 20Β A20\text{ A}, what is the current in the primary circuit?

Solution:

VpIp=VsIsV_p I_p = V_s I_s 2400Γ—Ip=120Γ—202400 \times I_p = 120 \times 20 Ip=24002400I_p = \frac{2400}{2400} Ip=1.0Β AI_p = 1.0\text{ A}

Explanation:

In an ideal transformer, power is conserved. We use the power equation VpIp=VsIsV_p I_p = V_s I_s to solve for the unknown primary current IpI_p.

Transformers - Revision Notes & Key Formulas | IGCSE Grade 11 Physics