Title: InkPhysics / Electricity
On every note of this deck, unless a note says otherwise:
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Front: DEFINITION
Define: Resistivity.
Back: The property of a material given by ρ = RA/L: the resistance of a piece of the material of unit length and unit cross-sectional area.

Front: DEFINITION
Define: Resistance.
Back: The ratio of the potential difference across a component to the current through it: R = V/I.

Front: CHECK YOURSELF
From a filament lamp's characteristic: at 2.0 V the current is 0.50 A, and at 12 V it is 1.5 A. Find the resistance at each point, and explain the change.
Show a hint
Resistance is a ratio of coordinates, taken twice.
Back: At 2.0 V: R = V/I = 2.0 / 0.50 = 4.0 Ω. At 12 V: R = 12 / 1.5 = 8.0 Ω.
The resistance has doubled because the larger current runs the filament far hotter: its ions vibrate more, the carriers collide with them more often, and the same pd drives proportionally less current.

Front: EQUATION — NOT ON THE DATA SHEET
Electrical energy transferred in time t:
E = ?
Back: E = IVt

Front: DEFINITION
Define: Semiconductor diode.
Back: A component that conducts with low resistance in the forward direction, above about 0.6 V, and has very high resistance in reverse.

Front: DEFINITION
Define: Electromotive force.
Back: The energy transferred to each coulomb of charge by a source: ε = E/Q, equal to the terminal pd when no current flows.

Front: DEFINITION
Define: Internal resistance.
Back: The resistance inside a source that makes the terminal pd fall below the emf when current flows: V = ε − Ir.

Front: DEFINITION
Define: Superconductor.
Back: A material whose resistivity falls to exactly zero when cooled below its critical temperature.

Front: DEFINITION
Define: Ohm's law.
Back: The current through an ohmic conductor is proportional to the potential difference across it, provided physical conditions such as temperature stay constant.

Front: CHECK YOURSELF
A divider is built from a 4.0 kΩ resistor above an 8.0 kΩ resistor across a 12 V supply, output across the 8.0 kΩ. (a) Find the output. (b) The 4.0 kΩ resistor is replaced by a thermistor, which then warms up. What happens to the output, and why?
Show a hint
Part (b) needs no numbers: follow the ratio.
Back: (a) V_out = 12 × 8.04.0 + 8.0 = 8.0 V.
(b) Warming an ntc thermistor lowers its resistance, so its share of the 12 V falls, and the share across the fixed 8.0 kΩ resistor, the output, rises towards 12 V. The supply never changed; the ratio did.

Front: DEFINITION
Define: Thermistor.
Back: A semiconductor component whose resistance falls steeply as temperature rises.

Front: EQUATION — NOT ON THE DATA SHEET
Potential divider output:
V_out = ?
Back: V_out = V_in × R_2R_1 + R_2

Front: CHECK YOURSELF
A wire of length 2.5 m and diameter 0.40 mm has a resistance of 1.2 Ω. Find the resistivity of its material.
Show a hint
Area from the diameter first, in metres, then rearrange the resistivity equation.
Back: Area: A = πd^2/4 = π × (0.40 × 10^−3)^2 / 4 = 1.26 × 10^−7 m^2.
ρ = RAL = 1.2 × 1.26 × 10^−7 / 2.5 = 6.0 × 10^−8 Ω m, a value typical of a pure metal.

Front: DEFINITION
Define: Potential difference.
Back: The work done per unit charge moving between two points: V = W/Q, in volts.

Front: CHECK YOURSELF
A 6.0 Ω and a 3.0 Ω resistor are connected in parallel, and the pair is in series with a 4.0 Ω resistor across a 12 V supply. Find the current drawn from the supply and the pd across the parallel pair.
Show a hint
Collapse the parallel pair first, then treat the circuit as a simple series loop.
Back: Parallel pair: 1R = 16.0 + 13.0 gives R = 2.0 Ω, smaller than either branch, as it must be.
Total resistance: 2.0 + 4.0 = 6.0 Ω, so the supply current is I = 12 / 6.0 = 2.0 A.
The pd across the pair: V = IR = 2.0 × 2.0 = 4.0 V, leaving 8.0 V across the 4.0 Ω resistor. The loop's 12 V is fully accounted for.

Front: DEFINITION
Define: Electric current.
Back: The rate of flow of charge: I = ΔQ/Δt.

Front: CHECK YOURSELF
A battery of emf 9.0 V and internal resistance 0.60 Ω drives a 4.4 Ω resistor. Find the current, the terminal pd, and the lost volts.
Show a hint
The emf pays for both resistances; the terminals only show the external share.
Back: Current: I = εR + r = 9.0 / (4.4 + 0.60) = 1.8 A.
Terminal pd: V = IR = 1.8 × 4.4 = 7.9 V (equivalently ε − Ir).
Lost volts: Ir = 1.8 × 0.60 = 1.1 V. The 9.0 V splits as 7.9 outside plus 1.1 inside, and the books balance.

Front: CHECK YOURSELF
A lamp carries a current of 0.25 A for 2.0 minutes. (a) How much charge passes through it? (b) How many electrons is that?
Show a hint
Seconds first. Then remember that charge comes in fixed lumps.
Back: (a) ΔQ = IΔt = 0.25 × 120 = 30 C.
(b) Each electron carries 1.60 × 10^−19 C, so n = 30 / 1.60 × 10^−19 = 1.9 × 10^20 electrons. A colossal number for a modest lamp; at counts like that, grainy charge behaves as a smooth fluid.