Notes
Current, resistance and Ohm's law
- Current $\displaystyle I = \frac{\Delta q}{\Delta t}$ (amperes). Conventional current flows from + to −.
- Resistance $\displaystyle R = \frac{\rho\ell}{A}$: longer, thinner wires and higher resistivity mean more resistance.
- Ohm's law: $\Delta V = IR$.
- Power: $\displaystyle P = I\Delta V = I^2R = \frac{(\Delta V)^2}{R}$.
Series and parallel
| Series | Parallel | |
|---|---|---|
| Current | Same through each | Splits among branches |
| Voltage | Splits among resistors | Same across each |
| Equivalent resistance | $R_{eq} = R_1 + R_2 + \dots$ | $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ |
- Adding a resistor in series increases $R_{eq}$; adding one in parallel decreases $R_{eq}$.
- Kirchhoff's junction rule: current in = current out (charge conservation). Loop rule: voltage changes around a loop add to zero (energy conservation).
- A real battery has internal resistance: terminal voltage $= \varepsilon - Ir$.
- Ammeters go in series; voltmeters go in parallel.
Capacitors in circuits
- $\displaystyle C = \frac{Q}{\Delta V}$; energy stored $\displaystyle U = \frac{1}{2}C(\Delta V)^2$.
- Parallel: $C_{eq} = C_1 + C_2$. Series: $\displaystyle \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
- RC circuits: when charging, current starts high and drops to zero as the capacitor fills. Time constant $\tau = RC$. After a long time, a capacitor acts like a break in the circuit.
Practice
$\displaystyle I = \frac{V}{R}$.
$\displaystyle \frac{1}{R} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$.
More resistance means less current through each bulb.
$\displaystyle I = \frac{P}{V}$.
Parallel branches connect to the same two points.
$\displaystyle R = \frac{\rho\ell}{A}$.
A fully charged capacitor blocks steady current.
Loop rule: $9.0 - 2.0 - 3.0 = 4.0$ V.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 12 V ideal battery is connected to a 4.0 Ω resistor in series with a parallel pair of 6.0 Ω and 12 Ω resistors.
(a) Find the equivalent resistance of the circuit.
Show the answer and scoring guide
Parallel pair: $\displaystyle \frac{1}{R} = \frac{1}{6} + \frac{1}{12} = \frac{1}{4}$, so 4.0 Ω. Total: $4.0 + 4.0 = 8.0$ Ω.
Check off the points you earned:
(b) Find the current through the battery and through the 6.0 Ω resistor.
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$\displaystyle I = \frac{12}{8.0} = 1.5$ A. Voltage across the parallel pair: $1.5 \times 4.0 = 6.0$ V, so $\displaystyle I_6 = \frac{6.0}{6.0} = 1.0$ A.
Check off the points you earned:
(c) The 12 Ω resistor is removed. Does the current through the 4.0 Ω resistor increase, decrease, or stay the same? Explain.
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Decrease. Removing a parallel branch raises the equivalent resistance (to 10 Ω), so the total current drops to 1.2 A.
Check off the points you earned: