Resistors and Capacitors

Resistors

• Conductors that have different resistivity
• Resistivity, $\rho = \frac{\left| \vec{E} \right| }{J} = R\frac{A}{l}$
• $R = \frac{\rho l}{A}$

Capcitors

• Most simply, plates that allow charges to distribute over their area.
• Capacitance, $C = \epsilon_0 \frac{A}{d}$

Loop Rule

$\Sigma V = 0$ Around any closed loop

Series Circuits

$V_{Batt}$ $- V_3$ $- V_2$ $- V_1 = 0$

We know V = IR, so

$V_{Batt}$ $- I_3R_3$ $- I_2R_2$ $- I_1R_1 = 0$

Series Circuits

$V_{Batt} - I_3R_3 - I_2R_2- I_1R_1 = 0$

I is the same everywhere in a series circuit

$V_{Batt} - I(R_3 + R_2 + R_1) = 0$

Series Circuits

$V_{Batt} - I(R_3 + R_2 + R_1) = 0$

In series circuits, then… $V_{Batt} - IR_{Total} = 0$

$R_{Total} = R_1 + R_2 + R_3+…$

Parallel Circuits

Choose a path $1 \rightarrow 8 \rightarrow 7 \rightarrow 2 \rightarrow 1$, then,

$V_{Batt} - I_1R_1 = 0$

Parallel Circuits

Choose a different path $1 \rightarrow 8 \rightarrow 6 \rightarrow 3 \rightarrow 1$

$V_{Batt} - I_2R_2 = 0$

Parallel Circuits

In each branch of a parallel circuit, $\Delta V$ must be the same.

$V_1 = V_2 = V_3 = V_{Batt}$

Parallel Circuits

In a parallel circuit, the current splits at each junction. So the current leaving the battery is $I_{Total} = I_1 + I_2 + I_3$

And, $V = IR$ or $I = V/R$ which can be substituted in.

$$I_{Total} = V_1/R_1 + V_2/R_2 + V_3/R_3$$

Parallel Circuits

$$I_{Total} = V_1/R_1 + V_2/R_2 + V_3/R_3$$ But, V is same across all parallel branches, so

$$V/R_{Total} = V/R_1 + V/R_2 + V/R_3$$ $$V/R_{Total} = V(1/R_1 + 1/R_2 + 1/R_3)$$

In Parallel branches $$1/R_{Total} = 1/R_1 + 1/R_2 + 1/R_3 + …$$

Series Circuits Current is same at all points $$ \Sigma \Delta V = 0$$ $R_{Total} = R_1 + R_2 + R_3+…$

Parallel Circuits Voltage is same across all branches $$I_{Total} = I_1 + I_2 + I_3$$ $1/R_{Total} = 1/R_1 + 1/R_2 + 1/R_3 + …$

Capacitors

Capacitors

As capacitor charges, charge builds up on plates of capacitor, creating electric field, building up a potential difference across the capacitor.

Capacitors

Charging Capacitor

Charge on capacitor: $Q = VC$

Capacitance: $C = \epsilon (A/d)$

Capacitors - Charging

Loop rule still applies, so

$V_{Batt}$ $-V_{Cap}$ $-V_{Res} = 0$

Capacitors - Charging

$$V_{Batt} -V_{Cap} -V_{Res} = 0$$

$$V_{Batt} -Q/C - IR = 0$$

$$V_{Batt} -Q(t)/C - \frac{dQ(t)}{dt}R = 0$$

Capacitors - Charging

$$V_{Batt} -Q(t)/C - \frac{dQ(t)}{dt}R = 0$$

• Look at the initial and final conditions.
  • At $t = 0, I = I_0, Q = Q_0, Q_0 often = 0 $
  • As $t \rightarrow \infty, I \rightarrow 0, Q \to VC$

$Q(t) = Q_0(1- e^{-t/RC})$

At time = 0 $$Q(t) = Q_0(1- e^{-t/RC})$$ $$\frac{dQ(t)}{dt} = \frac{Q_0}{RC}e^{-t/RC}$$ at t = 0, $Q(0) = Q_0$, and $I_0 = \frac{Q_0}{RC} = V/R$

As time goes to infinity $$Q(t) = Q_0(1- e^{-t/RC})$$ $$\frac{dQ(t)}{dt} = \frac{Q_0}{RC}e^{-t/RC}$$ as $t \to \infty, Q(\infty) = VC$, and $I(\infty) = I_0(1-1) = 0$

Capacitors - Disharging

Loop rule still applies, so

$V_{Cap}$ $-V_{Res} = 0$

Capacitors - Disharging

$$V_{Cap} -V_{Res} = 0$$

$$Q/C - IR = 0$$

$$Q(t)/C - \frac{dQ(t)}{dt}R = 0$$

Capacitors - Disharging

$$Q(t)/C - \frac{dQ(t)}{dt}R = 0$$

• Look at the initial and final conditions.
  • At $t = 0, I = I_0, Q_0 = VC$
  • As $t \to \infty, I \to 0, Q \to 0$

$Q(t) = VCe^{-t/RC}$

At time = 0 $$Q(t) = VCe^{-t/RC}$$ $$\frac{dQ(t)}{dt} = \frac{VC}{-RC}e^{-t/RC}$$ at t = 0, $Q(0) = VC$, and $I_0 = -\frac{VC}{RC} = -V_{C}/R$

As time goes to infinity $$Q(t) = VCe^{-t/RC}$$ $$\frac{dQ(t)}{dt} = \frac{VC}{-RC}e^{-t/RC}$$ as $t \to \infty, Q(\infty) = 0$, and $I(\infty) = V_{C}/R = 0$