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IAL Physics (Edexcel) · Topic 2.6 · 37 cards
Current, potential difference and resistance flashcards
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- Electric currentThe rate of flow of charged particles (rate of flow of charge). Unit: ampere, A.
- The coulombThe charge passing a point when a current of 1 A flows for 1 s: 1 C = 1 A s.
- Potential differenceThe energy transferred per unit charge between two points. Unit: volt, 1 V = 1 J C⁻¹.
- ResistanceR = V/I: the p.d. across a component divided by the current in it. Unit: ohm, 1 Ω = 1 V A⁻¹.
- Ohm's lawFor a metallic conductor at constant temperature, the current is directly proportional to the p.d. across it.
- Ohmic conductorA component whose resistance stays constant as the current changes: its I–V graph is a straight line through the origin.
- Drift velocityThe small average velocity of charge carriers along a conductor, superimposed on their random motion, when a p.d. is applied.
- Number density of charge carriers, nThe number of free charge carriers per unit volume (unit m⁻³).
- Conventional currentCurrent drawn from + to − round the external circuit: the direction positive charge would flow. Electrons in metals drift the opposite way.
- Current and chargeI = ΔQ/Δt (Yes)
- Potential difference, energy and chargeV = W/Q (Yes)
- Definition of resistanceR = V/I (Yes)
- Electrical power and energyP = VI and W = VIt (Yes)
- Power in terms of resistance (two forms)P = I²R and P = V²/R (Yes)
- Current in terms of drift velocityI = nqvA (Yes)
- Number of electrons in a charge QN = Q/e, with e = 1.60 × 10⁻¹⁹ C. (No — learn it)
- Area under a current–time graphThe charge that has passed.
- I–V graph of an ohmic conductorA straight line through the origin; resistance = V/I is the same at every point.
- Resistance from a curved I–V graphR = V/I using the coordinates of the point — not the gradient of the tangent.
- Shape of a P–V graph for a fixed resistorA parabola through the origin (P = V²/R); a smaller R gives a steeper curve.
- Where do the ammeter and voltmeter go?Ammeter in series with the component; voltmeter in parallel across it.
- Is current used up in a lamp?No. Charge is conserved, so current in = current out. Energy is transferred from the charge carriers to the lamp.
- Why does a lamp light at once although the drift velocity is tiny?The electric field is set up round the circuit almost instantly, so free electrons everywhere start drifting together.
- Why do electrons drift faster in a thinner part of a series circuit?Same I, n and q, so v ∝ 1/A: smaller area, larger drift velocity.
- Why do metals have a much lower resistivity than semiconductors and insulators?Metals have a huge n (~10²⁸–10²⁹ m⁻³); with fewer carriers, a given current needs a much larger drift velocity and so a much larger p.d.
- Why is Ohm's law only a special case of R = V/I?R = V/I defines resistance for any component; Ohm's law adds that R stays constant (metal at constant temperature).
- Same p.d., two resistors: which dissipates more power?The smaller resistance, because P = V²/R.
- A current of 0.60 A flows for 15 min. Charge, and number of electrons?Q = 0.60 × 900 = 540 C; N = 540 ÷ 1.60 × 10⁻¹⁹ = 3.4 × 10²¹.
- 25 C passes through a 9.0 V battery-powered toy. Energy transferred?W = QV = 25 × 9.0 = 225 J (2.3 × 10² J).
- A 1200 W hair dryer runs on 230 V. Current and resistance?I = P/V = 5.2 A; R = V²/P = 44 Ω.
- A 47 Ω resistor carries 0.30 A. Power dissipated?P = I²R = 0.30² × 47 = 4.2 W.
- A 60 W lamp is on 5 hours a day for 30 days. Energy in kWh?0.060 kW × 150 h = 9.0 kWh (3.2 × 10⁷ J).
- 3.0 A in a copper wire of area 2.5 mm², n = 8.5 × 10²⁸ m⁻³. Drift velocity?v = I/(nqA) = 3.0 ÷ (8.5 × 10²⁸ × 1.60 × 10⁻¹⁹ × 2.5 × 10⁻⁶) = 8.8 × 10⁻⁵ m s⁻¹.
- Stating Ohm's lawInclude ‘metallic conductor’, ‘constant temperature’ and ‘directly proportional’. V = IR alone is not the law.
- Converting mm² to m²× 10⁻⁶ (1 mm = 10⁻³ m, squared). Halve a diameter before squaring.
- Which V goes into P = V²/R?The p.d. across that same component — e.g. for a cable, IR across the cable, not the supply p.d.
- Current in a fixed resistor rises by 50%: percentage increase in power?Find the factor first (e.g. 1.5² = 2.25), then subtract 1: a 125% increase.