Series and parallel circuits and I–V characteristics flashcards

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Kirchhoff's first law
The sum of the currents into a junction equals the sum of the currents out of it.
Kirchhoff's second law
Around any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s.
Ohmic conductor
A conductor in which current is directly proportional to p.d. (I ∝ V) — for a metal, at constant temperature.
I–V characteristic
A graph of the current through a component against the p.d. across it, including negative (reversed) values.
Resistance at a point on an I–V graph
R = V/I, using the coordinates of that point — not the gradient.
Forward bias (diode)
The diode connected so that it conducts: current flows once the p.d. exceeds about 0.6 V (silicon).
Resistors in series
R = R₁ + R₂ + R₃ (Yes)
Resistors in parallel
1/R = 1/R₁ + 1/R₂ + 1/R₃ (Yes)
Two resistors in parallel (shortcut)
R = R₁R₂/(R₁ + R₂) — product over sum; two resistors only. (No — derive it)
n identical resistors R₀
Series: nR₀. Parallel: R₀/n. (No — derive it)
Currents at a junction
ΣI(in) = ΣI(out) (No — learn it)
I–V graph of an ohmic conductor
Straight line through the origin, the same in both directions; gradient = 1/R.
I–V graph of a filament lamp
Curve through the origin, symmetrical, with gradient decreasing as |V| increases (resistance rises as it heats).
I–V graph of an NTC thermistor
Curve through the origin, symmetrical, with gradient increasing as |V| increases (resistance falls as it heats).
I–V graph of a semiconductor diode
Almost zero current in reverse and below about 0.6 V forward; then current rises very steeply.
Circuit for an I–V characteristic
Variable resistor as a potential divider across the supply; ammeter in series with the component; voltmeter in parallel with the component.
Which I–V graph is not symmetrical about the origin?
The diode's — it conducts in one direction only.
Why is the current the same everywhere in a series circuit?
Charge is conserved and there are no junctions, so the charge passing each point per second is the same.
Why do series p.d.s add up to the supply p.d.?
Energy is conserved: each coulomb transfers in the components all the energy it gained in the source.
Why do parallel branches have the same p.d.?
Each branch joins the same two points; a coulomb takes one branch and transfers the same energy whichever it takes.
Why does adding a resistor in parallel reduce the total resistance?
It adds an extra path, so more current flows for the same p.d.; R = V/I falls below the smallest resistor.
Derive the series resistor equation
Same I in each; V = V₁ + V₂ + V₃ (energy); IR = IR₁ + IR₂ + IR₃; divide by I.
Derive the parallel resistor equation
Same V across each; I = I₁ + I₂ + I₃ (charge); V/R = V/R₁ + V/R₂ + V/R₃; divide by V.
Why use a potential divider, not a series variable resistor, for an I–V characteristic?
It lets the p.d. across the component be varied from zero, so the full graph near the origin can be plotted.
Why include a protective resistor when testing a diode?
Once the diode conducts its resistance is small; the resistor limits the current so the diode and meter are not damaged.
12 Ω, 18 Ω and 36 Ω in parallel. Combined resistance?
1/R = 3/36 + 2/36 + 1/36 = 6/36, so R = 6.0 Ω.
2.4 A and 0.9 A enter a junction; 1.5 A leaves along one wire. Current in the only other wire?
In = out: 2.4 + 0.9 = 1.5 + I, so I = 1.8 A (out).
6.0 V across 10 Ω in series with (20 Ω ∥ 30 Ω). Supply current and p.d. across the pair?
Pair = 12 Ω, total 22 Ω; I = 0.273 A; V(pair) = 0.2727 × 12 = 3.27 V.
12 V across 1.5 kΩ and 4.5 kΩ in series. P.d. across the 4.5 kΩ resistor?
Shared in ratio of resistance: 4.5/6.0 × 12 = 9.0 V.
A lamp's I–V graph passes through (6.0 V, 0.30 A). Resistance at 6.0 V?
R = V/I = 6.0 ÷ 0.30 = 20 Ω.
Diode (0.70 V) in series with 220 Ω on a 5.0 V supply. Current?
(5.0 − 0.70) ÷ 220 = 0.0195 A ≈ 20 mA.
What resistor in parallel with 220 Ω gives 150 Ω?
1/X = 1/150 − 1/220, X = 470 Ω (471 Ω).
You have added the reciprocals for a parallel network. What next?
Invert: the sum is 1/R. Then check R is smaller than the smallest resistor.
Combined network: which part first?
Replace each parallel group by its equivalent, add series parts, find the supply current, then work back out.
Two quick checks on any circuit answer
Loop p.d.s add up to the supply p.d.; branch currents add up to the current into the group.
Resistance from a curved I–V graph
Use V/I at the point. ΔV/ΔI (or 1/gradient) is not the resistance unless the line is straight through the origin.
Reverse-biased diode in series with a resistor: where is the supply p.d.?
Almost all across the diode — no current means no p.d. across the resistor.
Non-ohmic lamp in series with a resistor
Same current in both; test values from the I–V graph until V(lamp) + IR = supply p.d. Do not split the p.d. by resistance first.

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