B5-1. Charge, Voltage, Resistance, and Power

1. Quantisation of Charge and Electric Current

In electrical circuits, energy is transferred by the macroscopic movement of charge carriers. In metal wires, these carriers are negatively charged free electrons. Because charge is quantised, any total charge ($\Delta q$) is an integer multiple of the elementary charge ($e \approx 1.60 \times 10^{-19}$ C).

Definition of Electric Current ($I$):

Electric current is the rigorous time-rate of flow of electric charge across a specified cross-sectional boundary.

$$I = \dfrac{\Delta q}{\Delta t}$$
  • $I$: Current in Amperes (A). One Ampere is exactly one Coulomb of charge passing per second (C s⁻¹).
  • $\Delta q$: Net charge transferred in Coulombs (C).
  • $\Delta t$: Time interval in seconds (s).

Example 1: Calculating Macroscopic Charge Flow

Problem: A steady current of $4.5$ A flows through a copper conductor for $2.0$ minutes. Calculate the total charge transferred and the exact number of electrons that pass through the conductor's cross-section.

Solution:

  • Step 1: Calculate total charge ($\Delta q$).
    $\Delta t = 2.0 \times 60 = 120$ s
    $\Delta q = I \Delta t = (4.5) \times (120) = \mathbf{540 \text{ C}}$
  • Step 2: Relate to the elementary charge ($e$).
    $N = \dfrac{\Delta q}{e} = \dfrac{540}{1.60 \times 10^{-19}} = \mathbf{3.375 \times 10^{21} \text{ electrons}}$

2. Electric Potential Difference (Voltage)

Charge carriers do not flow through a macroscopic lattice spontaneously; work must be done on them to overcome resistive collisions. Potential difference is the continuous measure of this work.

Definition of Potential Difference ($V$):

The potential difference between two points is the electrical work done (or energy transferred) per unit of charge moving between those two points.

$$V = \dfrac{W}{q}$$
  • $V$: Potential difference in Volts (V). One Volt is equivalent to one Joule per Coulomb (J C⁻¹).
  • $W$: Work done or electrical energy transferred in Joules (J).
  • $q$: Charge moved in Coulombs (C).

Example 2: Work Done by a Power Supply

Problem: A power supply maintains a constant potential difference of $12.0$ V across a circuit. If a total of $3.0 \times 10^{20}$ electrons are pushed through the circuit, calculate the total electrical work done by the supply.

Solution:

  • Step 1: Calculate the total charge ($q$).
    $q = Ne = (3.0 \times 10^{20}) \times (1.60 \times 10^{-19}) = 48.0$ C
  • Step 2: Calculate the work done ($W$).
    $W = Vq = 12.0 \times 48.0 = \mathbf{576 \text{ J}}$

3. Resistance and Ohm's Law

Resistance ($R$) fundamentally defines how severely a component opposes current flow for a given potential difference. Ohm's Law is a specific physical condition stating that for ideal metallic conductors at a constant temperature, the current is strictly proportional to the voltage.

Definition of Resistance:

$$R = \dfrac{V}{I}$$
  • $R$: Resistance in Ohms (Ω), equivalent to Volts per Ampere (V A⁻¹).

Note: For non-Ohmic devices (like filament lamps), resistance increases at higher voltages because increased current causes lattice heating, exacerbating electron collisions.

Example 3: Dynamic Resistance in a Filament

Problem: A filament lamp draws $0.40$ A at $6.0$ V, but draws only $0.60$ A when the voltage is doubled to $12.0$ V. Calculate the resistance in both states and explain the physical meaning.

Solution:

  • State 1: $R_1 = \dfrac{6.0}{0.40} = \mathbf{15 \text{ Ω}}$
  • State 2: $R_2 = \dfrac{12.0}{0.60} = \mathbf{20 \text{ Ω}}$
  • Explanation: The lamp is non-Ohmic. The higher voltage dissipates more heat, raising the filament's temperature. The increased atomic vibration impedes electron flow more severely, increasing the absolute resistance.

4. Electrical Power and Joule Heating

Electrical power is the rate at which electrical energy is converted into other forms (such as thermal energy or mechanical work). By taking our foundational definitions ($I = q/t$ and $V = W/q$), we establish the base power equation: $P = \frac{W}{t} = \left(\frac{W}{q}\right)\left(\frac{q}{t}\right) = VI$.

Now, by substituting Ohm's Law ($V = IR$) into this base equation, we derive two specialized variations crucial for advanced circuit analysis.

Power Dissipation Equations:

$$P = VI = I^2 R = \dfrac{V^2}{R}$$
  • Use $P = I^2 R$ to analyze components in series (since current $I$ is constant everywhere).
  • Use $P = \dfrac{V^2}{R}$ to analyze components in parallel (since voltage $V$ is constant across branches).

Example 4: Optimizing Power Transmission

Problem: A power station transmits $800$ kW of electrical power to a city over transmission lines with a total resistance of $5.0$ Ω. Calculate the power completely lost as heat if the power is transmitted at $20,000$ V versus $100,000$ V.

Solution:

  • Case A ($20,000$ V):
    Transmission Current: $I = \dfrac{P}{V} = \dfrac{800,000}{20,000} = 40$ A.
    Power Lost (Heating): $P_{\text{loss}} = I^2R = (40)^2 \times 5.0 = 1600 \times 5.0 = \mathbf{8,000 \text{ W}}$
  • Case B ($100,000$ V):
    Transmission Current: $I = \dfrac{P}{V} = \dfrac{800,000}{100,000} = 8$ A.
    Power Lost (Heating): $P_{\text{loss}} = I^2R = (8)^2 \times 5.0 = 64 \times 5.0 = \mathbf{320 \text{ W}}$
  • Conclusion: Stepping up the voltage by a factor of 5 decreases the current by a factor of 5, which exponentially reduces the power lost by a factor of 25.