Flashcards · 9630 · AS Topic 4

Electricity (AS Level) flashcards

31 cards on AS Topic 4, subtopics 3.4.1 to 3.4.6. Pick a subtopic or study them all.

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Electric current: The rate of flow of charge: I = ΔQ/Δt. Unit A (C s⁻¹).
Potential difference: The work done (energy transferred) per unit charge moving between two points: V = W/Q. Unit V (J C⁻¹).
Resistance: The ratio of the p.d. across a component to the current through it: R = V/I. Unit Ω.
How many electrons pass a point each second when the current is 1.0 mA?: N = I/e = 1.0 × 10⁻³ ÷ 1.60 × 10⁻¹⁹ = 6.3 × 10¹⁵ per second.
Properties of an ideal ammeter and an ideal voltmeter: Ideal ammeter: zero resistance (connected in series). Ideal voltmeter: infinite resistance (connected in parallel).

Ohm's law: For a conductor, the current is proportional to the p.d. provided physical conditions such as temperature stay constant.
I-V characteristic of a filament lamp, and why: S-shaped curve through the origin, symmetrical, flattening at high p.d. As current rises the filament gets hotter, so its resistance increases.
I-V characteristic of a semiconductor diode: Almost no current in reverse bias, or in forward bias below about 0.6 V; above the threshold the current rises steeply. It conducts in one direction only.
Finding the resistance at a point on a curved I-V graph: Read the coordinates and calculate R = V/I. Do not use the gradient (or its reciprocal) of the curve.
Circuit for measuring an I-V characteristic: Ammeter in series with the component, voltmeter in parallel across it, and a variable supply (or potential divider) so the p.d. can be varied from near zero. Reverse the connections for negative values.

Resistivity: equation and unit: ρ = RA/L, so R = ρL/A. Unit Ω m. For a round wire A = πd²/4.
Finding resistivity from a graph of R against L: The gradient of the straight line through the origin is ρ/A, so ρ = gradient × A, with A from the measured diameter in metres.
Effect of temperature on the resistance of a metal and of an ntc thermistor: Metal: resistance increases, because the ions vibrate more and electrons collide with them more often. ntc thermistor: resistance decreases, because more charge carriers are released.
Superconductor and critical temperature: A superconductor has zero resistivity at and below its critical temperature.
Two uses of superconductors, and the drawback: Strong magnetic fields (scanners, particle accelerators) and power cables with no heating loss. The material must be kept below its critical temperature, which usually needs expensive cooling.
A wire's length and diameter are both doubled. What happens to its resistance?: L × 2 doubles R; A × 4 divides R by 4. Overall R halves.

Total resistance in series and in parallel: Series: R_T = R₁ + R₂ + R₃ + … Parallel: 1/R_T = 1/R₁ + 1/R₂ + 1/R₃ + …
Electrical power and energy equations: P = IV = I²R = V²/R. E = IVt.
Current and p.d. in series and in parallel circuits: Series: same current everywhere; the p.d.s add to the supply p.d. Parallel: same p.d. across each branch; the branch currents add to the total.
Which conservation laws apply in d.c. circuits?: Charge: current into a junction equals current out. Energy: round any loop, the sum of the emfs equals the sum of the p.d.s.
Combining identical cells in series and in parallel: Series: emfs add and internal resistances add. n identical cells in parallel: emf unchanged, internal resistance r/n.

Potential divider equation: V_out = V_in × R₂/(R₁ + R₂), where the output is taken across R₂.
ntc thermistor in a divider: what happens to the output across the fixed resistor as temperature rises?: The thermistor's resistance falls, so its share of the p.d. falls and the output across the fixed resistor rises.
LDR in a divider: output across the LDR as the light level rises: The LDR's resistance falls as light increases, so the p.d. across it falls.
Effect of connecting a load across the output of a potential divider: The load is in parallel with R₂, lowering the combined resistance, so the output p.d. falls. The smaller the load resistance, the bigger the drop.
Advantage of a potential divider over a variable resistor in series: A divider can vary the p.d. across the load from 0 V up to the supply p.d.; a series variable resistor cannot take it to zero.

Electromotive force (emf): The energy the source supplies per coulomb of charge it drives round the circuit: ε = E/Q.
Equations linking emf, terminal p.d. and internal resistance: ε = I(R + r), so terminal p.d. V = ε − Ir. Ir is the "lost volts".
Required practical: emf and internal resistance from a graph: Vary the external resistance and record terminal p.d. V and current I. Plot V against I: the vertical intercept is ε and the gradient is −r.
Short-circuit current of a cell: I = ε/r, the largest current the source can supply; it is why a shorted battery heats up.
ε = 1.80 V, r = 0.60 Ω, R = 2.4 Ω. Find the current and terminal p.d.: I = 1.80 ÷ 3.0 = 0.60 A. V = 0.60 × 2.4 = 1.44 V (lost volts 0.36 V).