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The Particulate Nature of Matter - Current and Circuits

Grade 12IBPhysics

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

🔑Concepts

Electric Current (II): The rate of flow of electric charge through a conductor, defined as I=ΔQΔtI = \frac{\Delta Q}{\Delta t}, measured in Amperes (AA).

Microscopic Model of Current: Current depends on the particulate nature of charge carriers. It is expressed as I=nAvqI = nAvq, where nn is the number density of charge carriers, AA is the cross-sectional area, vv is the drift velocity, and qq is the elementary charge (1.60×1019C1.60 \times 10^{-19} C).

Potential Difference (VV): The work done per unit charge in moving a small positive test charge between two points, V=WqV = \frac{W}{q}.

Ohm's Law: For many conductors (especially metals at constant temperature), the current is proportional to the potential difference, meaning resistance R=VIR = \frac{V}{I} remains constant.

Resistivity (ρ\rho): A property of the material itself, independent of geometry, defined by R=ρLAR = \rho \frac{L}{A}, where LL is length and AA is cross-sectional area.

Electromotive Force (EMF, ϵ\epsilon): The total energy supplied by a source per unit charge. Due to internal resistance (rr), the terminal potential difference (VV) is often less than the EMF: V=ϵIrV = \epsilon - Ir.

Kirchhoff's Laws: The Junction Law (conservation of charge) states Iin=Iout\sum I_{in} = \sum I_{out}. The Loop Law (conservation of energy) states ϵ=IR\sum \epsilon = \sum IR around any closed loop.

📐Formulae

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

I=nAvqI = nAvq

V=WqV = \frac{W}{q}

R=VIR = \frac{V}{I}

R=ρLAR = \rho \frac{L}{A}

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

ϵ=I(R+r)\epsilon = I(R + r)

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

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

💡Examples

Problem 1:

A copper wire has a cross-sectional area of 2.0×106m22.0 \times 10^{-6} m^2 and carries a current of 5.0A5.0 A. If the number density of free electrons in copper is 8.5×1028m38.5 \times 10^{28} m^{-3}, calculate the drift velocity of the electrons.

Solution:

Using the formula I=nAvqI = nAvq, we rearrange for vv: v=InAqv = \frac{I}{nAq} v=5.0(8.5×1028)(2.0×106)(1.6×1019)v = \frac{5.0}{(8.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})} v1.84×104ms1v \approx 1.84 \times 10^{-4} m s^{-1}

Explanation:

This demonstrates that while the signal of electricity travels near the speed of light, the actual 'particles' (electrons) move very slowly through the lattice of the conductor.

Problem 2:

A battery with an EMF of 12.0V12.0 V and an internal resistance of 0.5Ω0.5 \Omega is connected to a resistor of 5.5Ω5.5 \Omega. Determine the terminal potential difference of the battery.

Solution:

First, find the total current II using ϵ=I(R+r)\epsilon = I(R + r): I=ϵR+r=12.05.5+0.5=2.0AI = \frac{\epsilon}{R + r} = \frac{12.0}{5.5 + 0.5} = 2.0 A Now, find the terminal potential difference VV: V=ϵIr=12.0(2.0×0.5)=11.0VV = \epsilon - Ir = 12.0 - (2.0 \times 0.5) = 11.0 V

Explanation:

The terminal potential difference is lower than the EMF because some energy is dissipated as heat within the battery's internal resistance (IrIr 'lost volts').

Current and Circuits - Revision Notes & Key Formulas | IB Grade 12 Physics