Notes
Charge and Coulomb's law
- Like charges repel and opposite charges attract. Electron charge $= -1.6 \times 10^{-19} \text{ C}$. Charge is conserved.
- Conductors let charge move; insulators do not. Charge by friction, conduction or induction.
- $\displaystyle F = \frac{kq_1q_2}{r^2}$, $k = 9.0 \times 10^9 \text{ N·m}^2/\text{C}^2$; inverse-square.
Current, voltage and resistance
- Current $\displaystyle I = \frac{Q}{t}$ (amperes). Voltage $\displaystyle V = \frac{E}{Q}$ (volts). A closed circuit is needed.
- Ohm's law $V = IR$. Resistance grows with length and temperature and shrinks with thickness.
Series and parallel
| Series | Parallel | |
|---|---|---|
| current | same everywhere | splits among branches |
| voltage | splits among parts | same on every branch |
| total resistance | $R_1 + R_2 + \dots$ | $\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ |
Electric power
- $\displaystyle P = IV = I^2R = \frac{V^2}{R}$. Energy $E = Pt$; $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
The charged balloon pushes or pulls charges in the wall so the near side has the opposite charge, causing attraction.
Doubling $r$ divides $F$ by 4.
$\displaystyle R = \frac{V}{I} = \frac{5}{2} = 2.5 \text{ } \Omega$.
$\displaystyle \frac{1}{R} = \frac{3}{12}$, so $R = 4 \text{ } \Omega$.
There is only one path, so the same current flows through each part.
$\displaystyle I = \frac{P}{V} = \frac{60}{120} = 0.5 \text{ A}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 12 V battery is connected to a $4 \text{ } \Omega$ resistor in series with a parallel pair of resistors: one $6 \text{ } \Omega$ and one $12 \text{ } \Omega$.
(a) Find the total resistance of the circuit.
Show the answer and scoring guide
Parallel pair: $\displaystyle \frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12}$, so $R_p = 4 \text{ } \Omega$. Total $= 4 + 4 = 8 \text{ } \Omega$.
Check off the points you earned:
(b) Find the current from the battery and the voltage across the parallel pair.
Show the answer and scoring guide
$\displaystyle I = \frac{12}{8} = 1.5 \text{ A}$. Voltage across the $4 \text{ } \Omega$ series resistor $= 1.5 \times 4 = 6 \text{ V}$, so the parallel pair has $12 - 6 = 6 \text{ V}$.
Check off the points you earned:
(c) Find the current in the $6 \text{ } \Omega$ resistor and the power used by the $12 \text{ } \Omega$ resistor.
Show the answer and scoring guide
$\displaystyle I_6 = \frac{6}{6} = 1.0 \text{ A}$. For the $12 \text{ } \Omega$ resistor: $\displaystyle P = \frac{V^2}{R} = \frac{6^2}{12} = 3 \text{ W}$. (Check: $I_{12} = 0.5 \text{ A}$, and $1.0 + 0.5 = 1.5 \text{ A}$.)
Check off the points you earned: