Flashcards · 9702 · AS Topic 10

D.C. circuits (AS Level) flashcards

17 cards on AS Topic 10, subtopics 10.1 to 10.3. Pick a subtopic or study them all.

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Electromotive force (e.m.f.): The energy transferred per unit charge from other forms (such as chemical) to electrical, in driving charge round a complete circuit.
e.m.f. vs potential difference: e.m.f.: energy transferred to each unit charge by the source. p.d.: energy transferred from each unit charge into other forms in a component. Both are measured in volts.
Equation linking e.m.f., terminal p.d. and internal resistance: E = I(R + r) = V + Ir, so the terminal p.d. V = E − Ir. Ir is the "lost volts" inside the source.
Why does the terminal p.d. of a cell fall as the current increases?: More energy per unit charge is dissipated in the internal resistance (Ir increases), so less is available at the terminals. That is why headlights dim when a starter motor runs.
Terminal p.d. V plotted against current I: what do the gradient and intercept give?: Gradient = −r (the internal resistance). Intercept on the V axis = E (the e.m.f.).
Circuit symbols: cell and capacitor: Cell: two parallel lines of unequal length (long line = positive). Capacitor: two parallel lines of equal length.

Kirchhoff’s first law, and what it is a consequence of: The sum of the currents into a junction equals the sum of the currents out. A consequence of conservation of charge.
Kirchhoff’s second law, and what it is a consequence of: Around any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s. A consequence of conservation of energy.
Derive the formula for resistors in series: The same current I flows through each. By the second law, V = V₁ + V₂, so IR = IR₁ + IR₂, giving R = R₁ + R₂.
Derive the formula for resistors in parallel: Each has the same p.d. V. By the first law, I = I₁ + I₂, so V ÷ R = V ÷ R₁ + V ÷ R₂, giving 1/R = 1/R₁ + 1/R₂.
Find the combined resistance of 6.0 Ω and 3.0 Ω in parallel.: 1/R = 1/6.0 + 1/3.0 = 3/6.0, so R = 2.0 Ω, less than either resistor.
What happens to the total resistance when another resistor is added in parallel?: It decreases, because there is an extra path for current: more current flows for the same p.d.

Output of a potential divider: Vout = Vin × R₂ ÷ (R₁ + R₂), where R₂ is the resistor the output is taken across. The supply p.d. is shared in the ratio of the resistances.
12 V across a fixed 2.0 kΩ resistor and a 6.0 kΩ thermistor in series. Find the p.d. across the thermistor.: Vout = 12 × 6.0 ÷ (2.0 + 6.0) = 9.0 V.
The output is taken across an NTC thermistor in a potential divider. The temperature rises. What happens to Vout?: The thermistor’s resistance falls, so its share of the total resistance falls and Vout falls.
An LDR is in the lower position of a potential divider, with the output across it. It is illuminated. What happens to Vout?: The LDR’s resistance falls, so its share of the total resistance falls and Vout decreases.
Potentiometer and null method: how are two p.d.s compared?: Move the contact along the potentiometer until a galvanometer shows zero current (balance). No current is drawn from the p.d. being measured, so the measurement does not disturb the circuit.