Notes
Current and resistance
- $\displaystyle I = \frac{dq}{dt}$. Microscopically, $I = nqAv_d$ (drift velocity) and current density $\displaystyle J = \frac{I}{A} = \sigma E$.
- $\displaystyle R = \frac{\rho\ell}{A}$. Ohm's law $\Delta V = IR$ for ohmic materials. Power $P = I\Delta V = I^2R$.
- Real batteries: terminal voltage $= \varepsilon - Ir$.
Kirchhoff's rules
- Junction rule: $\displaystyle \sum I_{in} = \sum I_{out}$ (conservation of charge).
- Loop rule: $\displaystyle \sum\Delta V = 0$ around any closed loop (conservation of energy). Crossing a resistor in the direction of the current is a drop of $IR$.
- Set up one equation per independent loop and junction, then solve.
RC circuits
Charging through $R$ from a battery $\varepsilon$:
Discharging: $q(t) = Q_0e^{-t/RC}$.
- Time constant $\tau = RC$. After one $\tau$, a charging capacitor reaches about 63% of its final charge.
- Right after a switch closes, an uncharged capacitor acts like a wire. After a long time, it acts like a break (no current through it).
- Derive these from the loop rule: $\displaystyle \varepsilon - IR - \frac{q}{C} = 0$ with $\displaystyle I = \frac{dq}{dt}$.
Practice
$q = It$.
$12 - 2.0(1.0)$.
$\tau = (2000)(50 \times 10^{-6})$.
The uncharged capacitor acts like a wire.
$e^{-1} \approx 0.37$.
The total potential change around a loop must be zero.
Loop rule with $\displaystyle I = \frac{dq}{dt}$.
The capacitor is fully charged.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 12 V battery, a 3.0 kΩ resistor, a 2.0 μF uncharged capacitor and a switch are in series. The switch closes at $t = 0$.
(a) Find the current immediately after the switch closes.
Show the answer and scoring guide
$\displaystyle I_0 = \frac{12}{3000} = 4.0$ mA.
Check off the points you earned:
(b) Find the time constant and the charge on the capacitor at $t = \tau$.
Show the answer and scoring guide
$\tau = (3000)(2.0 \times 10^{-6}) = 6.0$ ms. $Q_{max} = C\varepsilon = 24$ μC. At $t = \tau$: $q = 24(1 - e^{-1}) \approx 15$ μC.
Check off the points you earned:
(c) Starting from the loop rule, derive $I(t)$.
Show the answer and scoring guide
$\displaystyle \varepsilon - IR - \frac{q}{C} = 0$. Differentiate: $\displaystyle -R\frac{dI}{dt} - \frac{I}{C} = 0 \Rightarrow \frac{dI}{I} = -\frac{dt}{RC} \Rightarrow I = \frac{\varepsilon}{R}e^{-t/RC}$.
Check off the points you earned: