Electric Current and Direct‑Current (D.C.) Circuits
Learning Objectives (Cambridge AS & A‑Level – Physics 9702)
- Define electric current as a flow of charge carriers and express it mathematically.
- Identify the charge carriers in metals, semiconductors, electrolytes and plasmas.
- Distinguish conventional current from electron flow.
- Explain potential difference, electrical power and the three power formulae.
- State Ohm’s law, define resistance and resistivity, and describe their temperature dependence.
- Interpret I‑V characteristics of resistors, filament lamps, diodes, light‑dependent resistors (LDRs) and thermistors.
- Analyse simple D.C. circuits using series/parallel rules, Kirchhoff’s laws and potential‑divider formulae.
- Apply practical techniques for measuring I, V and R and evaluate experimental uncertainties.
1. What is Electric Current?
Electric current (I) is the rate at which electric charge passes a given point in a circuit.
$$I=\frac{\Delta Q}{\Delta t}$$
- ΔQ – charge transferred (coulomb, C)
- Δt – time interval (second, s)
- 1 A = 1 C s⁻¹
2. Charge Carriers in Different Media
| Medium |
Charge Carrier(s) |
Typical Mobility (m² V⁻¹ s⁻¹) |
| Metals (e.g., Cu) |
Free electrons |
≈ 4.5 × 10⁻³ |
| n‑type semiconductor |
Electrons |
≈ 0.1 – 0.2 |
| p‑type semiconductor |
Holes |
≈ 0.05 – 0.1 |
| Electrolyte (e.g., NaCl solution) |
Positive & negative ions |
≈ 10⁻⁴ (ions) |
| Plasma |
Ions & free electrons |
Varies widely |
Conventional Current vs. Electron Flow
- Conventional current: direction a positive charge would move (from higher to lower potential).
- Electron flow: actual motion of electrons, opposite to conventional current in metals.
- Both conventions give the same magnitude of current; the conventional direction is retained for circuit analysis.
3. Potential Difference (Voltage) and Electrical Power
- Potential difference (ΔV): work done per unit charge to move a charge between two points
$$\Delta V=\frac{W}{Q}$$
Unit: volt (V) = joule per coulomb (J C⁻¹).
- Electrical power (P): rate at which electrical energy is transferred
$$P = VI = I^{2}R = \frac{V^{2}}{R}$$
Unit: watt (W) = joule per second (J s⁻¹).
Worked example – A 10 Ω resistor carries 0.25 A. Find the voltage across it and the power dissipated.
$$V = IR = (0.25\;\text{A})(10\;\Omega)=2.5\;\text{V}$$
$$P = VI = (2.5\;\text{V})(0.25\;\text{A})=0.625\;\text{W}$$
4. Resistance, Resistivity and Temperature Effects
- Resistance (R): opposition to the flow of charge; defined by Ohm’s law
$$V = IR$$
- Resistivity (ρ): intrinsic property of a material (Ω m). For a uniform conductor
$$R = \rho\,\frac{L}{A}$$
where L is length and A is cross‑sectional area.
- Temperature dependence (metals):
$$\rho(T)=\rho_{0}\,[1+\alpha\,(T-T_{0})]$$
α ≈ 3.9 × 10⁻³ K⁻¹ for copper.
I‑V Characteristics (syllabus requirement 9.3)
| Device |
Shape of I‑V curve |
Key features |
| Metallic resistor (Ohmic) |
Straight line through the origin |
Slope = 1/R, constant resistance. |
| Filament lamp (non‑Ohmic) |
Curve that steepens with increasing V |
Resistance rises as temperature rises. |
| Silicon diode |
Near‑zero current until V≈0.6 V (forward), then exponential rise; very small reverse current. |
Threshold (cut‑in) voltage ≈ 0.6 V. |
| Light‑dependent resistor (LDR) |
Resistance decreases sharply with increasing illumination; I‑V remains linear for a given light level. |
Photoconductivity – R ∝ 1/(light intensity). |
| Thermistor (NTC) |
Resistance falls exponentially with temperature; I‑V stays linear for a fixed temperature. |
R = R₀ e^{-βT} (β is material constant). |
5. Direct‑Current (D.C.) Circuits
5.1 Series and Parallel Combinations
| Configuration |
Current (I) |
Voltage (V) |
Equivalent Resistance (Req) |
| Series |
Same through all components |
Sum of individual drops |
Req = ΣRi |
| Parallel |
Same across each branch |
Sum of branch currents |
1/Req = Σ1/Ri |
5.2 Kirchhoff’s Laws
- Kirchhoff’s Current Law (KCL): The algebraic sum of currents entering a junction equals the sum leaving.
$$\sum I_{\text{in}} = \sum I_{\text{out}}$$
- Kirchhoff’s Voltage Law (KVL): The algebraic sum of potential differences round any closed loop is zero.
$$\sum V = 0$$
5.3 Potential Divider
Two series resistors R₁ and R₂ produce a fraction of the supply voltage:
$$V_{\text{out}} = V_{\text{supply}}\;\frac{R_{2}}{R_{1}+R_{2}}$$
Used for biasing semiconductor devices and for reference voltages.
5.4 Example Circuit Analysis
Find the current through each resistor in the series circuit shown (12 V battery, R₁=10 Ω, R₂=20 Ω, R₃=30 Ω).
- Total resistance: Req = 10 + 20 + 30 = 60 Ω.
- Total current: I = V/R = 12 V / 60 Ω = 0.20 A.
- Voltage across each resistor:
V₁ = IR₁ = 2 V, V₂ = 4 V, V₃ = 6 V.
6. Practical Skills (Paper 3 & Paper 5)
Measuring Current – Ammeter
- Connect in series with the component.
- Select the highest range first, then switch to a lower range for better resolution.
- Account for the ammeter’s internal resistance (usually negligible for digital meters).
- Uncertainty: ±(0.5 % of reading + 1 digit of the least‑significant place).
Measuring Voltage – Voltmeter
- Connect in parallel across the component.
- Use a high‑impedance voltmeter to minimise loading.
- Uncertainty similar to the ammeter; add systematic error from lead resistance for low voltages.
Measuring Resistance – Wheatstone Bridge / Ohmmeter
- For low resistances, employ the four‑wire (Kelvin) method to eliminate lead resistance.
- Record the balance length (or digital reading), calculate R, and propagate uncertainties.
Sample Experimental Question (Paper 5 style)
Design an experiment to verify Ohm’s law for a metal wire. Include a circuit diagram, list of apparatus, method of varying the voltage, how you will measure current and voltage, and how you will analyse the data (graph of V against I, gradient = R).
7. Summary
- Electric current is the flow of charge carriers; its magnitude is charge per unit time (A = C s⁻¹).
- Charge carriers differ by material: electrons in metals, electrons/holes in semiconductors, ions in electrolytes, both ions and electrons in plasmas.
- Conventional current direction is defined as the direction a positive charge would move; electron flow is opposite in metals.
- Potential difference (voltage) is work per unit charge; electrical power may be expressed as VI, I²R or V²/R.
- Resistance follows Ohm’s law (V = IR); resistivity links resistance to geometry and material (R = ρL/A) and varies with temperature.
- I‑V characteristics: linear for Ohmic resistors, non‑linear for filament lamps, diodes, LDRs and thermistors (see table).
- D.C. circuit analysis uses series/parallel rules, Kirchhoff’s laws and the potential‑divider formula.
- Practical measurements require correct instrument connections, appropriate range selection and careful uncertainty evaluation.
8. Practice Questions
- Calculate the current if $1.2\times10^{-2}\,\text{C}$ of charge passes a point in $3.0\,\text{s}$.
- A copper wire (density $8.96\times10^{3}\,\text{kg m}^{-3}$, atomic mass $63.5\,\text{g mol}^{-1}$) has a cross‑sectional area of $1.0\times10^{-6}\,\text{m}^{2}$ and an electron drift speed of $2.2\times10^{-4}\,\text{m s}^{-1}$. Estimate the current flowing through the wire.
- Explain why the direction of conventional current is opposite to the direction of electron flow in a metal circuit.
- Given a 12 V battery connected to a series circuit of three resistors (10 Ω, 20 Ω, 30 Ω), find the current through each resistor and the voltage across each.
- Sketch the I‑V characteristic of a silicon diode and label the forward‑bias threshold.
- Draw and analyse a circuit containing a 9 V battery, two loops and three resistors (R₁=5 Ω, R₂=10 Ω, R₃=15 Ω) using Kirchhoff’s laws. State the current in each branch.
- Design a simple experiment to verify the temperature dependence of a metal’s resistance. State the variables, method of measurement and the form of the graph you would plot.
- For an LDR, the resistance varies from $10 kΩ$ (dark) to $1 kΩ$ (bright). If a 5 V supply is applied across the LDR in series with a 1 kΩ resistor, calculate the voltage across the LDR in both lighting conditions.
- A thermistor (NTC) obeys $R = R_{0}e^{-\beta T}$ with $R_{0}=5 kΩ$ at $0^{\circ}\text{C}$ and $\beta = 0.02\;\text{K}^{-1}$. Find its resistance at $25^{\circ}\text{C}$.