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Physics: Electricity and Magnetism - DC Motors, Electromagnetic Induction, AC Generators, and Transformers

Grade 8IB

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

🔑Concepts

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The Motor Effect occurs when a current-carrying wire is placed in a magnetic field, experiencing a force. The direction of this force is determined by Fleming's Left Hand Rule, where the thumb represents Force (FF), the first finger represents Magnetic Field (BB), and the second finger represents Current (II).

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In a DC Motor, a split-ring commutator is used to reverse the direction of the current every half-turn, ensuring the motor continues to rotate in the same direction.

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Electromagnetic Induction is the production of an electromotive force (EMF) across an electrical conductor in a changing magnetic field. This is the operating principle for generators.

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Faraday's Law states that the magnitude of the induced EMF is directly proportional to the rate at which the magnetic flux linkage changes.

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Lenz's Law states that the direction of an induced current is always such as to oppose the change in magnetic flux that caused it (a consequence of the law of conservation of energy).

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An AC Generator uses slip rings and brushes to transfer the induced alternating current to an external circuit, converting mechanical energy into electrical energy.

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Transformers change the voltage of an alternating current using two coils (primary and secondary) wound around a soft iron core. A Step-up transformer increases voltage (Ns>NpN_s > N_p), while a Step-down transformer decreases voltage (Ns<NpN_s < N_p).

📐Formulae

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

VpIp=VsIsV_p I_p = V_s I_s

P=V×IP = V \times I

Efficiency=Useful Power OutTotal Power In×100%Efficiency = \frac{\text{Useful Power Out}}{\text{Total Power In}} \times 100\%

💡Examples

Problem 1:

A transformer is used to step down the mains voltage from 240240 V to 1212 V. If the primary coil has 40004000 turns, calculate the number of turns required in the secondary coil.

Solution:

Given: Vp=240V_p = 240 V, Vs=12V_s = 12 V, Np=4000N_p = 4000. Using the transformer equation: VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s} 24012=4000Ns\frac{240}{12} = \frac{4000}{N_s} 20=4000Ns20 = \frac{4000}{N_s} Ns=400020=200N_s = \frac{4000}{20} = 200

Explanation:

To decrease the voltage (Step-down), the number of turns in the secondary coil must be fewer than in the primary coil. In this case, 200200 turns are needed.

Problem 2:

An ideal transformer has a primary voltage of 220220 V and a primary current of 0.50.5 A. If the secondary voltage is 110110 V, calculate the secondary current.

Solution:

For an ideal transformer, input power equals output power: VpIp=VsIsV_p I_p = V_s I_s 220×0.5=110×Is220 \times 0.5 = 110 \times I_s 110=110×Is110 = 110 \times I_s Is=1 AI_s = 1 \text{ A}

Explanation:

Because the voltage is halved in this step-down transformer (220220 V to 110110 V), the current must double (0.50.5 A to 11 A) to maintain the same power level, assuming 100%100\% efficiency.

Problem 3:

Calculate the total turns in a primary coil if the secondary coil has 150150 turns, providing 3030 V from a 120120 V source.

Solution:

Using: VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s} 12030=Np150\frac{120}{30} = \frac{N_p}{150} 4=Np1504 = \frac{N_p}{150} 150×4600\begin{array}{r} 150 \\ \times 4 \\ \hline 600 \end{array} Np=600N_p = 600

Explanation:

The voltage ratio is 4:14:1, so the turns ratio must also be 4:14:1. Multiplying the secondary turns by 44 gives the primary turns.