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Physics - Electricity and Magnetism (Circuits and Induction)

Grade 10IB

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

🔑Concepts

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Electric Current (II): The rate of flow of electric charge (QQ) through a circuit, measured in Amperes (AA). I=QtI = \frac{Q}{t}.

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Potential Difference (VV): The work done (WW) per unit charge to move it between two points, measured in Volts (VV). V=WQV = \frac{W}{Q}.

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Ohm's Law: The current through a conductor between two points is directly proportional to the voltage across the two points, provided physical conditions like temperature remain constant (V=IRV = IR).

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Resistance in Series: The total resistance (RtotalR_{total}) is the sum of individual resistances. The current remains constant throughout the circuit.

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Resistance in Parallel: The reciprocal of the total resistance is the sum of the reciprocals of individual resistances. The potential difference remains constant across each branch.

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Electrical Power (PP): The rate at which electrical energy is converted into other forms, measured in Watts (WW).

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Electromagnetic Induction: The production of an electromotive force (emfemf) across an electrical conductor in a changing magnetic field.

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Faraday's Law: The magnitude of the induced emfemf is directly proportional to the rate of change of magnetic flux linkage.

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Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it.

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Transformers: Devices used to increase (step-up) or decrease (step-down) alternating voltages based on the ratio of turns in the primary (NpN_p) and secondary (NsN_s) coils.

📐Formulae

I=QtI = \frac{Q}{t}

V=IRV = IR

Rseries=R1+R2+R3+...R_{series} = R_1 + R_2 + R_3 + ...

1Rparallel=1R1+1R2+1R3+...\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...

P=VI=I2R=V2RP = VI = I^2R = \frac{V^2}{R}

E=Pt=VItE = Pt = VIt

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

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

💡Examples

Problem 1:

A circuit contains a 12V12 V battery connected to two resistors of 3Ω3 \Omega and 6Ω6 \Omega in parallel. Calculate the total current flowing from the battery.

Solution:

First, find total resistance (RpR_p): 1Rp=13+16=2+16=36=12Ω−1\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \Omega^{-1}. Thus, Rp=2ΩR_p = 2 \Omega. Now, use Ohm's Law: I=VRp=12V2Ω=6AI = \frac{V}{R_p} = \frac{12 V}{2 \Omega} = 6 A.

Explanation:

In parallel circuits, we sum the reciprocals of the resistances to find the equivalent resistance, then apply V=IRV = IR to find the total current.

Problem 2:

A step-down transformer has 40004000 turns on the primary coil and 200200 turns on the secondary coil. If the primary voltage is 240V240 V, calculate the secondary voltage.

Solution:

Using the transformer equation: VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}. Substituting the values: 240Vs=4000200\frac{240}{V_s} = \frac{4000}{200}. 240Vs=20⇒Vs=24020=12V\frac{240}{V_s} = 20 \Rightarrow V_s = \frac{240}{20} = 12 V.

Explanation:

The ratio of voltages is equal to the ratio of the number of turns in the coils.

Problem 3:

An electric heater is rated at 2kW,250V2 kW, 250 V. Calculate the resistance of the heating element when in use.

Solution:

Power P=2kW=2000WP = 2 kW = 2000 W. Using the formula P=V2RP = \frac{V^2}{R}, we rearrange to find RR: R=V2P=25022000=625002000=31.25ΩR = \frac{V^2}{P} = \frac{250^2}{2000} = \frac{62500}{2000} = 31.25 \Omega.

Explanation:

Resistance can be derived directly from the power rating and voltage using the algebraic rearrangement of P=VIP = VI and V=IRV = IR.

Electricity and Magnetism (Circuits and Induction) Revision - Grade 10 Science IB