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Electricity and Magnetism - Electric circuits

Grade 12IGCSEPhysics

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

πŸ”‘Concepts

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Electric Current (II): The rate of flow of electric charge, measured in Amperes (AA). Defined as I=dQdtI = \frac{dQ}{dt}.

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Electromotive Force (EMF, Ο΅\epsilon): The energy supplied by a source (like a battery) per unit charge to move it around a complete circuit.

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Potential Difference (VV): The energy transferred per unit charge between two points in a circuit, measured in Volts (VV).

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Ohm's Law: For an ohmic conductor, the current through it is directly proportional to the potential difference across it, provided physical conditions (like temperature) remain constant: V∝IV \propto I.

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Resistance (RR): The opposition to current flow. It depends on the material's resistivity (ρ\rho), length (LL), and cross-sectional area (AA): R=ρLAR = \frac{\rho L}{A}.

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Kirchhoff's First Law (Current Law): The sum of currents entering a junction is equal to the sum of currents leaving the junction (Conservation of Charge): βˆ‘Iin=βˆ‘Iout\sum I_{in} = \sum I_{out}.

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Kirchhoff's Second Law (Voltage Law): The sum of the EMFs in any closed loop is equal to the sum of the potential drops (βˆ‘IR\sum IR): βˆ‘Ο΅=βˆ‘IR\sum \epsilon = \sum IR.

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Internal Resistance (rr): The inherent resistance within a power source that causes a drop in terminal potential difference when current flows.

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Potential Dividers: Circuits used to provide a specific output voltage (VoutV_{out}) by splitting the source voltage between two or more resistors.

πŸ“Formulae

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

V=WQV = \frac{W}{Q}

V=IRV = IR

R=ρLAR = \frac{\rho L}{A}

Rtotal=R1+R2+R3+...Β (Series)R_{total} = R_1 + R_2 + R_3 + ... \text{ (Series)}

1Rtotal=1R1+1R2+1R3+...Β (Parallel)\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... \text{ (Parallel)}

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

Ο΅=I(R+r)\epsilon = I(R + r) or Ο΅=Vterminal+Ir\epsilon = V_{terminal} + Ir

Vout=(R2R1+R2)VinV_{out} = \left( \frac{R_2}{R_1 + R_2} \right) V_{in}

πŸ’‘Examples

Problem 1:

A battery with an EMF of 12.0V12.0 V and an internal resistance of 0.5Ξ©0.5 \Omega is connected to an external resistor of 5.5Ξ©5.5 \Omega. Calculate the terminal potential difference across the battery.

Solution:

First, find the total resistance: Rtotal=R+r=5.5Ξ©+0.5Ξ©=6.0Ξ©R_{total} = R + r = 5.5 \Omega + 0.5 \Omega = 6.0 \Omega. Next, calculate the circuit current: I=Ο΅Rtotal=12.0V6.0Ξ©=2.0AI = \frac{\epsilon}{R_{total}} = \frac{12.0 V}{6.0 \Omega} = 2.0 A. Finally, calculate the terminal voltage: V=Ο΅βˆ’Ir=12.0Vβˆ’(2.0AΓ—0.5Ξ©)=11.0VV = \epsilon - Ir = 12.0 V - (2.0 A \times 0.5 \Omega) = 11.0 V.

Explanation:

The terminal potential difference is always less than the EMF when a current flows because of the voltage drop across the internal resistance (IrIr).

Problem 2:

Calculate the total resistance of three resistors (2Ξ©2 \Omega, 4Ξ©4 \Omega, and 4Ξ©4 \Omega) connected in parallel.

Solution:

Using the parallel resistance formula: 1Rtotal=12+14+14\frac{1}{R_{total}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{4}. This simplifies to 1Rtotal=24+14+14=44=1Ξ©βˆ’1\frac{1}{R_{total}} = \frac{2}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4} = 1 \Omega^{-1}. Therefore, Rtotal=1Ξ©R_{total} = 1 \Omega.

Explanation:

In a parallel circuit, the total resistance is always less than the smallest individual resistance.

Problem 3:

A 2.0kW2.0 kW electric kettle is connected to a 250V250 V mains supply. Calculate the current flowing through the element and its resistance.

Solution:

Using P=VIP = VI, current I=PV=2000W250V=8.0AI = \frac{P}{V} = \frac{2000 W}{250 V} = 8.0 A. Using V=IRV = IR, resistance R=VI=250V8.0A=31.25Ξ©R = \frac{V}{I} = \frac{250 V}{8.0 A} = 31.25 \Omega.

Explanation:

Power must be converted to Watts (1kW=1000W1 kW = 1000 W) before using the standard formulae.

Electric circuits - Revision Notes & Key Formulas | IGCSE Grade 12 Physics