Creado por Niamh Walsh
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Pregunta | Respuesta |
Physics Unit 2: ELECTRICITY | KEY TOPICS: - current and potential difference - power and work -resistance and resistivity - potential dividers and internal resistance |
Current: rate of flow of charged particles I = ∆Q/∆t Potential difference (voltage): the energy transfer between two points V = W/Q (overleaf: IV graphs) | |
Resistance: a measure of the opposition a material offers the flow of current. The unit is Ohms (Ω) R = V/I | Ohm's Law The current is directly proportional to the voltage across a wire, provided the physical factors remain constant |
Resistance increases with temperature. Most metals are conductors - there are free electrons to carry a charge. | A semi-conductor contains far fewer free electrons. - negative temperature coefficient: resistance ⬇️ temperature ⬆️ - positive temperature coefficient resistance ⬆️ temperature ⬇️ |
Materials have a constant resistivity, which can be found using R = ⍴l/A where resistivity is ⍴ | |
Power: the rate of work done P = ∆W/∆t P = VQ/t P = VI Work done W = VIt | In a series circuit: current is constant, voltage decreases around the circuit In a parallel circuit: current splits at each junction, voltage is constant |
Kirchhoff's First Law The total current into any point in a circuit is equal to the total current out of that point | Kirchhoff's Second Law The sum of potential difference which rises and falls around a closed path in a circuit is zero |
Number density of conduction electrons n = I / qvA n is large - good conductor n is small - bad conductor | Voltage = V Current = I (amp) Resistance = R (Ohms) Work done = W Power = P Resistivity = ⍴ Q = charge (coulombs) v = drift velocity A = cross-sectional area l = length n = number of free electrons t = time r = internal resistance E = e.m.f. |
The Potential Divider Two resistors connected in series (so current same, p.d. divided between the two) with a supply. V(1) = R(!) ___ ___ V(2) R(2) | Internal resistance Electro Motive Force: total energy the power supply supplies each unit of charge e.m.f. = V + Ir Ir is often known as the 'lost volts' 1 eV = 1.6 x10^-19 J |
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