IB — physicsMedium2 min read

Current, Voltage, and Ohm's Law

I = ΔQ/Δt; V = W/Q; Ohm's law V = IR; I-V characteristics; resistance and resistivity

Tutor AI Team
Tutor AI Team

# Current, Voltage, and Ohm's Law — IB Physics

Electrical circuits are analysed using current, voltage, and resistance. Ohm's law provides the fundamental relationship, while I-V characteristics reveal the behaviour of different components.

1. Current

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

1 ampere = 1 coulomb per second. Conventional current: positive to negative (opposite to electron flow).

2. Potential Difference (Voltage)

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

1 volt = 1 joule per coulomb.

EMF (ε\varepsilon): energy per unit charge from a source. PD: energy per unit charge used by a component.

3. Resistance and Ohm's Law

V=IR\boxed{V = IR}

Ohm's Law: For an ohmic conductor at constant temperature, current is proportional to voltage.

R=ρLAR = \frac{\rho L}{A}

Resistivity (ρ\rho) is a material property.

4. I-V Characteristics

Component Curve Notes
Ohmic resistor Straight line through origin RR constant
Filament lamp Curve (flattening) RR increases with TT
Diode Near-zero current until ~0.7V Very low RR in forward bias

5. EMF and Internal Resistance

ε=I(R+r)\varepsilon = I(R + r) V=ε−IrV = \varepsilon - Ir

"Lost volts" = IrIr. Terminal PD < EMF when current flows.

Worked Example: Example 1

Problem

A 12 V battery with 0.5 Ω internal resistance drives a 5.5 Ω external resistor. I=12/(5.5+0.5)=2I = 12/(5.5 + 0.5) = 2 A Vterminal=12−2(0.5)=11V_{\text{terminal}} = 12 - 2(0.5) = 11 V

Solution

Worked Example: Example 2

Problem

A copper wire (ρ=1.7×10−8\rho = 1.7 \times 10^{-8} Ω·m) of length 10 m and diameter 1 mm. A=π(0.5×10−3)2=7.85×10−7A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m² R=1.7×10−8×10/7.85×10−7=0.217R = 1.7 \times 10^{-8} \times 10/7.85 \times 10^{-7} = 0.217 Ω

Solution

7. Practice Questions

    1. Define EMF and internal resistance. (2 marks)
    1. A battery of EMF 9V and internal resistance 1Ω is connected to a 4Ω resistor. Find current and terminal PD. (3 marks)
    1. Sketch I-V graphs for an ohmic resistor and a filament lamp. Explain the difference. (4 marks)

    Answers

    1. EMF: energy per unit charge converted from other forms to electrical by the source. Internal resistance: resistance of the source itself.
    1. I=9/(4+1)=1.8I = 9/(4+1) = 1.8 A. V=9−1.8(1)=7.2V = 9 - 1.8(1) = 7.2 V.

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Summary

  • I=ΔQ/ΔtI = \Delta Q/\Delta t; V=W/QV = W/Q; V=IRV = IR
  • R=ρL/AR = \rho L/A
  • EMF: ε=I(R+r)\varepsilon = I(R+r); terminal PD = ε−Ir\varepsilon - Ir
  • Ohmic: straight I-V; filament: curved (R increases with T)
Tags:
#ib#physics#electricity#current#voltage#ohms-law#resistance

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