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Electricity and Magnetism - Electrical quantities

Grade 12IGCSEPhysics

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

🔑Concepts

Electric Charge (QQ): The fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is measured in Coulombs (CC). The charge of a single electron is approximately 1.6imes1019C-1.6 imes 10^{-19} C.

Electric Current (II): The rate of flow of electric charge. It is defined as I=QtI = \frac{Q}{t} and measured in Amperes (AA), where 1A=1C/s1 A = 1 C/s. Conventional current flows from positive to negative, whereas electrons flow from negative to positive.

Potential Difference (VV): The work done (WW) per unit charge (QQ) in moving a charge between two points in a circuit. It is measured in Volts (VV), where 1V=1J/C1 V = 1 J/C.

Electromotive Force (e.m.f.): The total energy supplied by a source (like a battery) per unit charge as it moves around a complete circuit. It is also measured in Volts (VV).

Resistance (RR): The opposition to the flow of electric current. According to Ohm's Law, for an ohmic conductor at constant temperature, the current is directly proportional to the potential difference, leading to V=IRV = IR. Resistance is measured in Ohms (Ω\Omega).

Resistivity (ρ\rho): An intrinsic property of a material that quantifies how strongly it opposes the flow of electric current. Resistance RR depends on resistivity, length (LL), and cross-sectional area (AA).

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

📐Formulae

Q=ItQ = I \cdot t

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

V=IRV = I \cdot R

R=ρLAR = \frac{\rho \cdot L}{A}

P=VIP = V \cdot I

P=I2RP = I^2 \cdot R

P=V2RP = \frac{V^2}{R}

W=VItW = V \cdot I \cdot t

💡Examples

Problem 1:

A copper wire has a length of 2.0m2.0 m and a cross-sectional area of 5.0×107m25.0 \times 10^{-7} m^2. If the resistivity of copper is 1.7×108Ωm1.7 \times 10^{-8} \Omega \cdot m, calculate the resistance of the wire.

Solution:

R=ρLA=(1.7×108Ωm)×(2.0m)5.0×107m2=0.068ΩR = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8} \Omega \cdot m) \times (2.0 m)}{5.0 \times 10^{-7} m^2} = 0.068 \Omega

Explanation:

To find the resistance, we use the resistivity formula. By substituting the given length (LL), area (AA), and resistivity (ρ\rho), we find that the total resistance is 0.068Ω0.068 \Omega.

Problem 2:

A light bulb is connected to a 230V230 V power supply. If the current flowing through the bulb is 0.5A0.5 A, calculate the power rating of the bulb and the energy consumed in 1010 minutes.

Solution:

Power: P=VI=230V×0.5A=115WP = V \cdot I = 230 V \times 0.5 A = 115 W\nEnergy: W=Pt=115W×(10×60s)=69,000J=69kJW = P \cdot t = 115 W \times (10 \times 60 s) = 69,000 J = 69 kJ

Explanation:

First, we calculate power using P=VIP = VI. Then, to find the energy, we multiply power by time, ensuring time is converted from minutes to seconds (10min=600s10 min = 600 s).

Problem 3:

A charge of 120C120 C flows through a resistor in 22 minutes. Calculate the current in the circuit.

Solution:

I=Qt=120C2×60s=1.0AI = \frac{Q}{t} = \frac{120 C}{2 \times 60 s} = 1.0 A

Explanation:

Current is defined as the rate of flow of charge. By dividing the total charge (120C120 C) by the time in seconds (120s120 s), we find the current is 1.0A1.0 A.

Electrical quantities - Revision Notes & Key Formulas | IGCSE Grade 12 Physics