Kirchhoff's Laws, Potential Dividers, Thermistors/LDRs, EMF & Internal Resistance E = I(R + r)
Direct current (DC) circuit analysis relies on two fundamental conservation laws formulated by Kirchhoff: conservation of electric charge at any junction (ΣI_in = ΣI_out) and conservation of energy around any closed loop (Σε = ΣIR). While metallic ohmic conductors maintain a constant resistance at constant temperature, heating a metal filament lamp increases lattice ion vibrations (raising resistance), whereas heating an NTC semiconductor thermistor or illuminating an LDR liberates mobile charge carriers into the conduction band (exponentially lowering resistance). Real power supplies possess internal resistance r, causing the terminal potential difference V to fall linearly with current I according to V = -rI + ε. Examiners heavily test sensor potential divider circuits, resistivity (R = ρL / A), and graphical determination of EMF ε and internal resistance r.
- Current (I = ΔQ / Δt), Potential Difference (V = W / Q) & Resistivity (ρ = RA / L)
- Electric current I is the rate of flow of charge (I = ΔQ / Δt, in Amperes A = C s⁻¹). Potential difference V is the work done (electrical energy transferred to other forms) per unit charge between two points (V = W / Q, in Volts V = J C⁻¹). Resistance is defined as the ratio R = V / I (never the gradient of an I-V curve unless it is a straight line through the origin). Resistivity ρ is an intrinsic material property: R = ρL / A, or ρ = RA / L (units: Ω m).
- I-V Characteristics: Ohmic Resistor, Filament Lamp, NTC Thermistor & Diode
- 1. Ohmic conductor (metal wire at constant T): I is directly proportional to V (straight line through origin). 2. Filament lamp: as V and I increase, the metal filament heats up; positive metal ions vibrate with greater amplitude, increasing collision frequency with conduction electrons, so resistance R increases (I-V curve flattens). 3. NTC (Negative Temperature Coefficient) Thermistor & LDR: made of intrinsic semiconductors; increasing temperature (or light intensity on an LDR) excites electrons into the conduction band, vastly increasing number density n of charge carriers so resistance R decreases sharply (I-V curve bends steeply upwards). 4. Semiconductor Diode: conducts only in forward bias above a threshold voltage (~0.6-0.7 V for silicon); almost infinite resistance in reverse bias.
- Kirchhoff's First & Second Laws, Series/Parallel Resistors & Power
- 1. Kirchhoff's First Law (Conservation of Charge): the sum of currents entering any junction equals the sum of currents leaving that junction (ΣI_in = ΣI_out). 2. Kirchhoff's Second Law (Conservation of Energy): around any closed loop in a circuit, the algebraic sum of the EMFs equals the algebraic sum of the potential drops (Σε = ΣIR). Series resistors: R_total = R₁ + R₂ + . (same current I). Parallel resistors: 1 / R_total = 1 / R₁ + 1 / R₂ + . (same p.d. V). Electrical Power: P = VI = I²R = V² / R, and Energy E = VIt.
Electric Field Simulator & Interactive NPN Transistor Builder
Adjust charges and distances to visualize Coulomb’s Law field lines, or switch to the Transistor Lab to wire and test an NPN circuit.
1. Coulomb’s Law Point-Charge Calculation
Formula: F = k · |Q₁ · Q₂| / r² where k = 1/(4πε₀) = 8.99 × 10⁹ N·m²·C⁻²
3.60 N
Attractive (towards each other)7.19e+6 N/C
At r/2 = 0.10 m2. Uniform Parallel-Plate Field Calculator
Formula: E = V / d (units: V m⁻¹ or N C⁻¹)
Key Concepts & Exam Definitions (6)
- Current (I = ΔQ / Δt), Potential Difference (V = W / Q) & Resistivity (ρ = RA / L)
- Electric current I is the rate of flow of charge (I = ΔQ / Δt, in Amperes A = C s⁻¹). Potential difference V is the work done (electrical energy transferred to other forms) per unit charge between two points (V = W / Q, in Volts V = J C⁻¹). Resistance is defined as the ratio R = V / I (never the gradient of an I-V curve unless it is a straight line through the origin). Resistivity ρ is an intrinsic material property: R = ρL / A, or ρ = RA / L (units: Ω m).
- I-V Characteristics: Ohmic Resistor, Filament Lamp, NTC Thermistor & Diode
- 1. Ohmic conductor (metal wire at constant T): I is directly proportional to V (straight line through origin). 2. Filament lamp: as V and I increase, the metal filament heats up; positive metal ions vibrate with greater amplitude, increasing collision frequency with conduction electrons, so resistance R increases (I-V curve flattens). 3. NTC (Negative Temperature Coefficient) Thermistor & LDR: made of intrinsic semiconductors; increasing temperature (or light intensity on an LDR) excites electrons into the conduction band, vastly increasing number density n of charge carriers so resistance R decreases sharply (I-V curve bends steeply upwards). 4. Semiconductor Diode: conducts only in forward bias above a threshold voltage (~0.6-0.7 V for silicon); almost infinite resistance in reverse bias.
- Kirchhoff's First & Second Laws, Series/Parallel Resistors & Power
- 1. Kirchhoff's First Law (Conservation of Charge): the sum of currents entering any junction equals the sum of currents leaving that junction (ΣI_in = ΣI_out). 2. Kirchhoff's Second Law (Conservation of Energy): around any closed loop in a circuit, the algebraic sum of the EMFs equals the algebraic sum of the potential drops (Σε = ΣIR). Series resistors: R_total = R₁ + R₂ + . (same current I). Parallel resistors: 1 / R_total = 1 / R₁ + 1 / R₂ + . (same p.d. V). Electrical Power: P = VI = I²R = V² / R, and Energy E = VIt.
- Potential Divider Circuits with Thermistors & LDRs
- Two resistors R₁ and R₂ in series across a supply V_in divide the voltage in direct proportion to their resistances: V_out (across R₂) = V_in × [R₂ / (R₁ + R₂)], and V₁ / V₂ = R₁ / R₂. Golden Rule for Sensor Questions (4-step chain): 1. State how the environmental change (e.g. darker or colder) affects the resistance of the LDR or NTC thermistor (R_sensor increases). 2. Therefore, the sensor takes a larger share (proportion) of the total supply voltage V_in. 3. Therefore, V_across_sensor increases (while V_across_fixed_resistor = V_in - V_sensor decreases). 4. Relate this to switching on a heater, fan, or street light.
- Electromotive Force (ε), Internal Resistance (r) & Lost Volts (Ir)
- Electromotive force (EMF, ε = E / Q) is the chemical/other energy converted into electrical energy per unit charge by the source (equal to the open-circuit terminal p.d. when zero current flows, I = 0). Every real cell has an internal resistance r due to resistance of its chemicals/electrodes. By conservation of energy: ε = I(R + r) = V + Ir, where V = IR is the terminal potential difference across the external load R, and v = Ir is the 'lost volts' (energy wasted as heat per coulomb inside the cell).
- Graphical Determination of EMF (ε) and Internal Resistance (r)
- Connect an ammeter in series and a high-resistance voltmeter across the terminals of a cell in series with a variable resistor R. Vary R and record pairs of V and I. Rearranging ε = V + Ir into linear y = mx + c form gives: V = (-r)I + ε. Plotting Terminal p.d. V (y-axis) against Current I (x-axis) produces a straight line with a negative gradient: 1. The y-intercept (at I = 0) equals the EMF ε. 2. The gradient of the line equals -r (so internal resistance r = -gradient). 3. The x-intercept (at V = 0) is the short-circuit current I_max = ε / r.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Explain at the microscopic level why the resistance of a metal wire increases with temperature, whereas the resistance of an NTC thermistor decreases with temperature.
State what physical conservation law underpins Kirchhoff's First Law and Kirchhoff's Second Law.
A wire of resistance R = 12 Ω is melted down and recast into a new uniform wire of half the original length (½L) with no loss of metal. What is the resistance of the new wire?
In a potential divider with a fixed resistor R₁ and an LDR in series across a 9.0 V battery, across which component should the output voltmeter V_out be connected so that V_out increases when it gets dark?
Define Electromotive Force (EMF, ε) and explain why the terminal p.d. V of a battery drops below ε when a current I is drawn.
On a graph of Terminal Potential Difference V (y-axis) against Current I (x-axis) for a cell, how do you find the EMF ε, internal resistance r, and short-circuit current?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A nichrome heating wire has a length of 1.80 m, a diameter of 0.36 mm, and a resistance of 19.5 Ω at 20°C. Calculate the resistivity of nichrome at 20°C (stating the correct SI unit), and explain how and why the current decreases slightly during the first few seconds after the wire is connected to a constant 12.0 V DC power supply. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate radius r = 0.18 × 10⁻³ m and cross-sectional area A = πr² = π × (0.18 × 10⁻³)² = 1.018 × 10⁻⁷ m² [1 mark]. 2. Rearrange R = ρL / A for resistivity ρ: ρ = RA ÷ L = (19.5 × 1.018 × 10⁻⁷) ÷ 1.80 = 1.10 × 10⁻⁶ [1 mark]. 3. State unit of resistivity: Ω m [1 mark]. 4. When connected to the 12.0 V supply, current flows and electrical energy is dissipated as thermal energy (P = I²R), causing the temperature of the nichrome wire to rise rapidly [1 mark]. 5. As temperature increases, the positive metal lattice ions vibrate with greater amplitude, increasing the frequency of collisions with delocalised conduction electrons; this increases the resistance R of the wire and reduces the current I = V / R [1 mark].
- Q2
A battery of EMF ε and internal resistance r is connected in series with a variable resistor R. When R is set to 4.0 Ω, the current in the circuit is 1.50 A. When R is increased to 9.0 Ω, the current falls to 0.75 A. Calculate the internal resistance r and the EMF ε of the battery, and calculate the power wasted inside the battery when R = 4.0 Ω. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Use ε = I(R + r) to write two simultaneous equations: Equation 1: ε = 1.50(4.0 + r) = 6.0 + 1.50r [1 mark]. 2. Equation 2: ε = 0.75(9.0 + r) = 6.75 + 0.75r [1 mark]. 3. Equate the two expressions for ε: 6.0 + 1.50r = 6.75 + 0.75r → 0.75r = 0.75 → r = 1.0 Ω [1 mark]. 4. Substitute r = 1.0 Ω back into Equation 1: ε = 1.50(4.0 + 1.0) = 7.5 V [1 mark]. 5. Power wasted inside the battery when I = 1.50 A: P_lost = I²r = (1.50)² × 1.0 = 2.25 W [1 mark].
- Q3
In a temperature-sensing potential divider circuit, a 12.0 V battery of negligible internal resistance is connected in series with a fixed resistor of resistance R₁ = 2.40 kΩ and an NTC thermistor. At 20°C, the thermistor has a resistance of 3.60 kΩ. A high-resistance voltmeter is connected across the fixed resistor R₁. Calculate the reading on the voltmeter at 20°C, and explain what happens to the voltmeter reading when the thermistor is placed in hot water at 80°C. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Use the potential divider equation across R₁: V_out = V_in × [R₁ ÷ (R₁ + R_thermistor)] = 12.0 × [2.40 ÷ (2.40 + 3.60)] [1 mark]. 2. Calculate V_out at 20°C: V_out = 12.0 × (2.40 ÷ 6.00) = 4.80 V [1 mark]. 3. When placed in hot water at 80°C, thermal energy liberates more charge carriers in the NTC semiconductor thermistor, so its resistance R_thermistor decreases significantly [1 mark]. 4. Because R_thermistor takes a smaller share of the 12.0 V supply p.d. (and total circuit resistance falls so current I increases), the potential difference across the fixed resistor R₁ (V_out = IR₁) increases [1 mark].
- Q4
In the potential divider circuit above (12.0 V supply, R₁ = 2.40 kΩ, R_thermistor = 3.60 kΩ at 20°C), a student replaces the ideal digital voltmeter across R₁ with an analogue moving-coil voltmeter that has a finite resistance of 4.80 kΩ. Calculate the new reading on the analogue voltmeter at 20°C. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. The 4.80 kΩ voltmeter is connected in parallel with the 2.40 kΩ fixed resistor R₁ [1 mark]. 2. Calculate the combined resistance R_p of the parallel combination: 1 / R_p = 1 / 2.40 + 1 / 4.80 = 3 / 4.80 → R_p = 4.80 ÷ 3 = 1.60 kΩ [1 mark]. 3. The circuit now acts as a potential divider between R_p (1.60 kΩ) and R_thermistor (3.60 kΩ), with total resistance = 1.60 + 3.60 = 5.20 kΩ [1 mark]. 4. Calculate the new voltmeter reading: V_out = 12.0 × (1.60 ÷ 5.20) = 3.69 V (lower than 4.80 V because the non-ideal voltmeter draws current) [1 mark].
Common Exam Questions (4)
Why does a car battery have a very low internal resistance (~0.01 Ω), whereas an EHT high-voltage lab power supply (5000 V) has a huge internal resistance (~5 MΩ) built in?
A car starter motor requires a massive current (~200 A) from a 12 V battery to crank the engine; since I_max = ε / r, internal resistance r must be tiny (~0.01 Ω) or all 12 V would be lost inside the battery. Conversely, a 5000 V school EHT supply has a huge safety resistor r = 5 MΩ in series inside it so that even if a student touches the terminals (short circuit), the maximum current is limited to I_max = 5000 ÷ 5×10⁶ = 1 mA, preventing fatal electrocution.
At what value of external load resistance R is the maximum electrical power transferred from a battery of internal resistance r to the external load?
By the Maximum Power Transfer Theorem, maximum power is delivered to the external load when the external resistance exactly matches the internal resistance of the source (R = r). At R = r, the efficiency is 50% and P_max = ε² / (4r).
What is a superconductor, and what are its practical applications?
A superconductor is a material whose resistivity drops suddenly to exactly zero (0 Ω m) when cooled below its critical transition temperature (T_c). Once a current is induced in a superconducting loop, it flows indefinitely with zero I²R heat loss (V = 0), making superconductors ideal for generating intense magnetic fields in hospital MRI scanners, CERN particle accelerators (LHC), and Maglev trains.
When reading an I-V graph for a non-ohmic component like a filament bulb or diode at a specific voltage V₁, how do I find the resistance R at that point?
Never take the gradient (ΔI / ΔV) of the tangent to the curve. Resistance is always defined as the simple coordinate ratio R = V₁ / I₁ read directly off the axes at that point.
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