Capacitance and DC Circuits

Big Picture

A capacitor stores charge and energy in an electric field. Current is the flow of charge. Resistance limits current. Together, capacitors, resistors, and voltage sources form DC circuits. Kirchhoff's junction rule (charge conservation) and loop rule (energy conservation) are the universal tools for circuit analysis.

ELI-10: Explain It Like I'm 10

A capacitor is like a tiny rechargeable water balloon for electricity — it fills up with charge and holds it until needed. Current is how fast electric charge flows through a wire, like water flowing through a pipe. Resistance is how much a material fights against that flow — a thin wire has more resistance, like a narrow pipe. Circuits are paths that electricity flows through; switches and components control where it goes.

11.1 Capacitance

Core Idea

Capacitance measures a device's ability to store charge for a given voltage.

\[ C = \frac{Q}{V} \]

SI unit: farad (F). \(1\ \text{F} = 1\ \text{C/V}\). Most practical capacitors are in μF (10⁻⁶ F) or pF (10⁻¹² F).

Parallel-plate capacitor:

\[ C = \epsilon_0\frac{A}{d} \]

Here \(\epsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}\) is the permittivity of free space, \(A\) is the plate area, and \(d\) is the plate separation. This formula tells us: larger plates store more charge (bigger \(A\)), and plates closer together store more charge (smaller \(d\)) — because the electric field between them is stronger.

With a dielectric (insulating material between plates):

\[ C = \kappa\epsilon_0\frac{A}{d} = \kappa C_0 \]

where \(\kappa\) is the dielectric constant (dimensionless, always > 1). The dielectric does two things: (1) it reduces the electric field between the plates because the dielectric's molecules polarize and create an opposing field, and (2) it allows the plates to be placed closer together without sparking. Both effects increase capacitance. Common dielectric constants: vacuum \(\kappa = 1\), paper \(\kappa \approx 3.5\), glass \(\kappa \approx 5\)–\(10\), strontium titanate \(\kappa \approx 300\).

Energy stored in a capacitor:

\[ U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV \]

This energy is stored in the electric field between the plates. The energy density (energy per unit volume) in the field is:

\[ u = \frac{1}{2}\epsilon_0 E^2 \]

Worked Example: Parallel-Plate Capacitor with Dielectric

Problem: A parallel-plate capacitor has square plates of side length 10 cm separated by 1.0 mm of air. Find (a) its capacitance, (b) the charge stored when connected to a 12 V battery, (c) the new capacitance if the air gap is filled with mica (\(\kappa = 6.0\)), and (d) the energy stored before and after inserting the dielectric (battery remains connected).

Solution:

(a) Plate area: \(A = (0.10\ \text{m})^2 = 0.010\ \text{m}^2\). Plate separation: \(d = 1.0 \times 10^{-3}\ \text{m}\).

\[ C_0 = \epsilon_0\frac{A}{d} = (8.85 \times 10^{-12})\frac{0.010}{1.0 \times 10^{-3}} = 8.85 \times 10^{-11}\ \text{F} = 88.5\ \text{pF} \]

(b) Charge with air gap:

\[ Q_0 = C_0 V = (8.85 \times 10^{-11})(12) = 1.06 \times 10^{-9}\ \text{C} = 1.06\ \text{nC} \]

(c) With mica dielectric (\(\kappa = 6.0\)), and the battery still connected so \(V\) stays at 12 V:

\[ C = \kappa C_0 = 6.0 \times 88.5\ \text{pF} = 531\ \text{pF} \]

The capacitance increases sixfold because the dielectric reduces the internal field, allowing more charge to accumulate for the same voltage.

(d) Energy with air gap:

\[ U_0 = \frac{1}{2}C_0 V^2 = \frac{1}{2}(8.85 \times 10^{-11})(12)^2 = 6.37 \times 10^{-9}\ \text{J} \]

Energy with mica (battery connected, \(V\) constant):

\[ U = \frac{1}{2}CV^2 = \frac{1}{2}(5.31 \times 10^{-10})(12)^2 = 3.82 \times 10^{-8}\ \text{J} \]

The stored energy increases by a factor of \(\kappa = 6.0\) because \(U \propto C\) when \(V\) is constant. The additional energy comes from the battery, which does work to push more charge onto the plates against the increased capacitance.

Key insight: If the battery were disconnected before inserting the dielectric (so \(Q\) stays constant), the energy would decrease by a factor of \(\kappa\). The dielectric would be pulled into the gap by the fringe field — mechanical work is done.

Capacitor Energy: Worked Example — Camera Flash

Problem: A camera flash unit uses a 470 μF capacitor charged to 300 V. Find (a) the energy stored, (b) the average power if this energy is discharged in 2.0 ms, and (c) the charge stored on the capacitor.

Solution:

(a) Energy stored:

\[ U = \frac{1}{2}CV^2 = \frac{1}{2}(470 \times 10^{-6})(300)^2 = 21.15\ \text{J} \]

(b) Average power during the flash:

\[ P_{\text{avg}} = \frac{U}{\Delta t} = \frac{21.15}{2.0 \times 10^{-3}} = 1.06 \times 10^4\ \text{W} \approx 10.6\ \text{kW} \]

This is why a tiny capacitor can produce a blinding flash — it releases energy extremely quickly, achieving enormous instantaneous power despite storing only a modest amount of energy.

(c) Charge stored:

\[ Q = CV = (470 \times 10^{-6})(300) = 0.141\ \text{C} \]

ELI-10: Explain It Like I'm 10

A capacitor is two metal plates separated by a small gap. When you connect a battery, electrons pile up on one plate (negative) and are pulled away from the other (positive). The plates hold this charge even after you disconnect the battery — like a tiny rechargeable battery. The wider the plates and the closer they are, the more charge they store. The stored energy can be released quickly (camera flash) or used to smooth out voltage (power supplies).

11.2 Electric Current and Ohm's Law

Core Ideas

Electric current \(I = \Delta Q/\Delta t\). SI unit: ampere (A). \(1\ \text{A} = 1\ \text{C/s}\).

Conventional current flows from + to − (historical convention). Electron flow is opposite — electrons drift from − to +.

Drift Velocity

Although the electric signal propagates through a wire at nearly the speed of light, individual electrons move surprisingly slowly. The drift velocity \(v_d\) is the average velocity of charge carriers in response to an electric field, and it is typically on the order of mm/s.

The drift velocity is given by:

\[ v_d = \frac{I}{nqA} \]

where:

• \(I\) is the current (A)
• \(n\) is the number density of charge carriers (m⁻³)
• \(q\) is the charge per carrier (\(1.60 \times 10^{-19}\ \text{C}\) for electrons)
• \(A\) is the cross-sectional area of the wire (m²)

Worked estimate: Consider a 1.0 A current in a copper wire of diameter 1.0 mm. Copper has \(n \approx 8.5 \times 10^{28}\ \text{m}^{-3}\) free electrons per cubic meter. The cross-sectional area:

\[ A = \pi r^2 = \pi(5.0 \times 10^{-4})^2 = 7.85 \times 10^{-7}\ \text{m}^2 \]

\[ v_d = \frac{1.0}{(8.5 \times 10^{28})(1.60 \times 10^{-19})(7.85 \times 10^{-7})} \approx 9.3 \times 10^{-5}\ \text{m/s} \approx 0.1\ \text{mm/s} \]

That is roughly 0.3 meters per hour — barely crawling. Yet when you flip a light switch, the light turns on instantly. Why? Because the electric field propagates at near light speed, pushing all electrons in the circuit almost simultaneously, like water already filling a pipe — when you turn on the tap at one end, water immediately comes out the other end, even though individual water molecules take a long time to travel the length of the pipe.

Ohm's Law

For ohmic materials (constant resistance):

\[ V = IR \]

Resistance \(R\). SI unit: ohm (Ω). \(1\ \Omega = 1\ \text{V/A}\).

Resistance of a wire: \(R = \rho\frac{L}{A}\), where \(\rho\) is resistivity (Ω·m). Longer wire → more resistance. Thicker wire → less resistance.

Not all materials are ohmic. Filament light bulbs, diodes, and semiconductors have nonlinear \(I\)–\(V\) curves. In an ohmic resistor, a graph of \(I\) vs. \(V\) is a straight line through the origin; the slope is \(1/R\). In a non-ohmic device, the graph curves.

Resistivity and Temperature

For most metals, resistivity increases linearly with temperature over moderate ranges:

\[ \rho = \rho_0[1 + \alpha(T - T_0)] \]

where \(\alpha\) is the temperature coefficient of resistivity. For copper, \(\alpha \approx 3.9 \times 10^{-3}\ \text{°C}^{-1}\). This is why a light bulb's filament has much lower resistance when cold — the initial current surge when turned on can be 10–15 times the steady operating current, which is why bulbs most often burn out at the moment they are switched on.

Electrical Power

Electrical power: \(P = IV = I^2R = V^2/R\). SI unit: watt (W). \(1\ \text{W} = 1\ \text{J/s}\).

The form \(P = I^2R\) represents power dissipated as heat in a resistor (Joule heating). The form \(P = V^2/R\) is useful when voltage is known directly.

Worked Example: Electrical Power

Problem: An electric heater is rated at 1500 W when connected to 120 V. Find (a) its resistance, (b) the current it draws, and (c) the cost to run it for 8.0 hours if electricity costs $0.12 per kWh.

Solution:

(a) Using \(P = V^2/R\):

\[ R = \frac{V^2}{P} = \frac{(120)^2}{1500} = 9.6\ \Omega \]

(b) Current:

\[ I = \frac{P}{V} = \frac{1500}{120} = 12.5\ \text{A} \]

(Check: \(V = IR = 12.5 \times 9.6 = 120\ \text{V}\) ✓)

(c) Energy used: \(E = P \times t\). Convert to kWh:

\[ E = 1.50\ \text{kW} \times 8.0\ \text{h} = 12.0\ \text{kWh} \]

\[ \text{Cost} = 12.0 \times \$0.12 = \$1.44 \]

ELI-10: Explain It Like I'm 10

Current is the flow rate of electric charge — how many coulombs pass through each second. Voltage is the "push" that drives them. Resistance is how much the wire resists the flow. Ohm's law says: current = voltage ÷ resistance. More push or less resistance = more current. A light bulb converts electrical power (\(P = IV\)) into light and heat.

11.3 Resistors in Series and Parallel

Core Ideas

Series (same current through all):

\[ R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots \]

In a series circuit, the total voltage divides across the resistors, with the largest voltage drop across the largest resistance. The current is the same everywhere.

Parallel (same voltage across all):

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \]

In a parallel circuit, the total current divides among the branches, with the largest current flowing through the smallest resistance. The voltage is the same across every branch.

For two resistors in parallel, a convenient shortcut:

\[ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \]

The equivalent resistance of a parallel combination is always less than the smallest individual resistance.

Worked Example: Series–Parallel Combination

Problem: Three resistors — \(R_1 = 6.0\ \Omega\), \(R_2 = 3.0\ \Omega\), \(R_3 = 2.0\ \Omega\) — are connected to a 12 V battery. \(R_2\) and \(R_3\) are in parallel, and this parallel pair is in series with \(R_1\). Find (a) the equivalent resistance, (b) the total current, (c) the current through each resistor, and (d) the voltage across each resistor.

Solution:

(a) First, the parallel combination of \(R_2\) and \(R_3\):

\[ R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{(3.0)(2.0)}{3.0 + 2.0} = \frac{6.0}{5.0} = 1.2\ \Omega \]

Now \(R_{23}\) is in series with \(R_1\):

\[ R_{\text{eq}} = R1 + R{23} = 6.0 + 1.2 = 7.2\ \Omega \]

(b) Total current from the battery:

\[ I{\text{total}} = \frac{V}{R{\text{eq}}} = \frac{12}{7.2} = 1.67\ \text{A} \]

(c) This total current flows through \(R_1\) (series): \(I_1 = 1.67\ \text{A}\). The voltage across \(R_1\) is:

\[ V_1 = I_1 R_1 = (1.67)(6.0) = 10.0\ \text{V} \]

The voltage across the parallel pair is what remains:

\[ V_{23} = V - V_1 = 12 - 10.0 = 2.0\ \text{V} \]

Now, both \(R_2\) and \(R_3\) have 2.0 V across them (parallel):

\[ I2 = \frac{V{23}}{R_2} = \frac{2.0}{3.0} = 0.667\ \text{A} \]

\[ I3 = \frac{V{23}}{R_3} = \frac{2.0}{2.0} = 1.0\ \text{A} \]

Check: \(I_2 + I_3 = 0.667 + 1.0 = 1.667\ \text{A} = I_1\) ✓ (junction rule).

(d) Summary:

Capacitors: Opposite Rules

Series: \(\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots\) (reduces total capacitance).

Parallel: \(C_{\text{eq}} = C_1 + C_2 + \cdots\) (increases total capacitance).

The rules for capacitors are the reverse of resistors because capacitance is proportional to plate area (which adds in parallel) but inversely proportional to plate separation (which adds in series). Physically, capacitors in series have the same charge on each — the equivalent capacitance is smaller because the effective plate separation increases.

ELI-10: Explain It Like I'm 10

Resistors in series are like narrow pipes connected end to end — the total resistance adds up. In parallel, they are like multiple pipes side by side — more paths mean less total resistance. Capacitors are the opposite: parallel adds their storage capacity; series reduces it.

11.4 Kirchhoff's Rules

Core Ideas

Junction Rule: \(\sum I{\text{in}} = \sum I{\text{out}}\). Charge is conserved — current does not disappear at a junction. Think of it as: "what flows in must flow out."

Loop Rule: \(\sum \Delta V = 0\) around any closed loop. Energy is conserved — the total voltage gained equals the total voltage dropped. If you trace a path around any closed loop, you return to the same potential you started at.

Sign conventions:

• Going through a battery from − to +: +V (voltage gain).
• Going through a battery from + to −: −V (voltage drop).
• Going through a resistor in the direction of current: −IR (voltage drop).
• Going through a resistor opposite the direction of current: +IR (voltage gain).

Worked Example: Multi-Loop Circuit

Problem: A circuit has two batteries and three resistors. The left loop contains a 12 V battery and resistors \(R_1 = 4.0\ \Omega\) and \(R_2 = 6.0\ \Omega\). The right loop contains a 6.0 V battery (polarity opposing the 12 V battery) and \(R_2\) (shared) and \(R_3 = 3.0\ \Omega\). Find the current in each branch.

Circuit description:

• Left branch: 12 V battery (positive terminal up) in series with \(R_1 = 4.0\ \Omega\).
• Middle branch: \(R_2 = 6.0\ \Omega\).
• Right branch: 6.0 V battery (negative terminal up) in series with \(R_3 = 3.0\ \Omega\).
• All three branches connect at top and bottom junctions.

Solution:

Step 1: Label currents. Let \(I_1\) flow up through the left branch (through the 12 V battery and \(R_1\)), \(I_2\) flow down through the middle branch (\(R_2\)), and \(I_3\) flow up through the right branch (through \(R_3\) and the 6.0 V battery).

Step 2: Junction rule at the top junction:

\[ I_1 = I_2 + I_3 \quad \text{(1)} \]

Step 3: Loop rule — left loop (clockwise, starting at bottom left):

\[ +12 - I_1 R_1 - I_2 R_2 = 0 \quad \text{(2)} \]

\[ 12 - 4.0 I_1 - 6.0 I_2 = 0 \]

Step 4: Loop rule — right loop (clockwise, starting at bottom right):

Going clockwise: up through \(R_3\), hit the 6.0 V battery from − to + (gain), then down through \(R_2\):

\[ -I_3 R_3 + 6.0 + I_2 R_2 = 0 \quad \text{(3)} \]

\[ -3.0 I_3 + 6.0 + 6.0 I_2 = 0 \]

Step 5: Solve the system. From (1): \(I_3 = I_1 - I_2\). Substitute into (3):

\[ -3.0(I_1 - I_2) + 6.0 + 6.0 I_2 = 0 \]

\[ -3.0 I_1 + 3.0 I_2 + 6.0 + 6.0 I_2 = 0 \]

\[ -3.0 I_1 + 9.0 I_2 + 6.0 = 0 \quad \text{(4)} \]

From (2): \(4.0 I_1 + 6.0 I_2 = 12\), so \(I_1 = 3.0 - 1.5 I_2\).

Substitute into (4):

\[ -3.0(3.0 - 1.5 I_2) + 9.0 I_2 + 6.0 = 0 \]

\[ -9.0 + 4.5 I_2 + 9.0 I_2 + 6.0 = 0 \]

\[ 13.5 I_2 - 3.0 = 0 \quad \Rightarrow \quad I_2 = 0.222\ \text{A} \]

Then \(I_1 = 3.0 - 1.5(0.222) = 3.0 - 0.333 = 2.67\ \text{A}\).

And \(I_3 = I_1 - I_2 = 2.67 - 0.222 = 2.45\ \text{A}\).

Step 6: Verify — check the right loop:

\[ -3.0(2.45) + 6.0 + 6.0(0.222) = -7.35 + 6.0 + 1.33 = -0.02 \approx 0\ \text{✓} \]

Interpretation: The 12 V battery (stronger) drives 2.67 A through the left branch. At the top junction, 0.222 A goes down through \(R_2\) and 2.45 A goes through the right branch. The 6.0 V battery is being charged (current flows opposite to its polarity) — this is how a car battery charges from the alternator.

ELI-10: Explain It Like I'm 10

Kirchhoff's two rules are just common sense: (1) The water flowing into a pipe junction equals the water flowing out — nothing disappears. (2) If you walk around a closed loop, the total uphill climbs equal the total downhill descents — you end up where you started energetically.

11.5 RC Circuits

Core Idea

An RC circuit contains a resistor and capacitor. The voltage and current change exponentially during charging and discharging. Unlike pure resistive circuits (which reach steady state instantly), RC circuits have a characteristic time delay.

Time constant \(\tau = RC\) (seconds). It measures how quickly the capacitor charges or discharges.

• After \(t = \tau\), the capacitor is ~63% charged (or discharged).
• After \(t = 2\tau\): ~86%.
• After \(t = 3\tau\): ~95%.
• After \(t = 5\tau\): >99% — considered "fully" charged/discharged for practical purposes.

Charging a capacitor (series RC, switch closed at \(t = 0\)):

\[ q(t) = CV(1 - e^{-t/RC}) \]

\[ I(t) = \frac{V}{R} e^{-t/RC} \]

\[ V_C(t) = V(1 - e^{-t/RC}) \]

At \(t = 0\), the uncharged capacitor acts like a short circuit (zero voltage across it, maximum current). At \(t \to \infty\), the capacitor acts like an open circuit (full voltage, zero current).

Discharging a capacitor (series RC, switch closed at \(t = 0\), initial charge \(Q_0\)):

\[ q(t) = Q_0 e^{-t/RC} \]

\[ I(t) = -\frac{Q_0}{RC} e^{-t/RC} = -I_0 e^{-t/RC} \]

\[ V_C(t) = V_0 e^{-t/RC} \]

The negative sign in the current indicates the direction is opposite to the charging current.

Worked Example 1: RC Charging

Problem: A 5.0 μF capacitor charges through a 2.0 kΩ resistor from a 9.0 V battery. Find (a) the time constant, (b) the charge after one time constant, (c) the time to reach 90% of full charge, and (d) the initial current.

Solution:

(a) \(\tau = RC = (2000)(5.0\times 10^{-6}) = 0.010\ \text{s} = 10\ \text{ms}\).

(b) After \(t = \tau\):

\[ q = CV(1 - e^{-1}) = (5.0\times 10^{-6})(9.0)(0.632) = 2.84 \times 10^{-5}\ \text{C} = 28.4\ \mu\text{C} \]

(c) We need \(q = 0.90 \times CV = 0.90 \times 45\ \mu\text{C} = 40.5\ \mu\text{C}\).

\[ 0.90 = 1 - e^{-t/\tau} \quad \Rightarrow \quad e^{-t/\tau} = 0.10 \]

\[ -\frac{t}{\tau} = \ln(0.10) = -2.303 \quad \Rightarrow \quad t = 2.303\tau = 2.303 \times 10\ \text{ms} = 23\ \text{ms} \]

(d) Initial current (at \(t = 0\), capacitor uncharged):

\[ I_0 = \frac{V}{R} = \frac{9.0}{2000} = 4.5\ \text{mA} \]

After one time constant: \(I = I_0 e^{-1} = 4.5 \times 0.368 = 1.66\ \text{mA}\).

Worked Example 2: RC Discharging

Problem: The same 5.0 μF capacitor, fully charged to 9.0 V, is disconnected from the battery and connected across the 2.0 kΩ resistor. Find (a) the initial discharge current, (b) the voltage across the capacitor after 15 ms, and (c) the time for the voltage to drop to 1.0 V.

Solution:

(a) Initial discharge current (magnitude):

\[ I_0 = \frac{V_0}{R} = \frac{9.0}{2000} = 4.5\ \text{mA} \]

(b) After \(t = 15\ \text{ms} = 1.5\tau\):

\[ V_C = V_0 e^{-t/\tau} = 9.0 \, e^{-1.5} = 9.0 \times 0.223 = 2.01\ \text{V} \]

(c) We need \(V_C = 1.0\ \text{V}\):

\[ 1.0 = 9.0\, e^{-t/\tau} \quad \Rightarrow \quad e^{-t/\tau} = \frac{1.0}{9.0} = 0.111 \]

\[ -\frac{t}{\tau} = \ln(0.111) = -2.197 \quad \Rightarrow \quad t = 2.197\tau = 2.197 \times 10\ \text{ms} = 22\ \text{ms} \]

ELI-10: Explain It Like I'm 10

A capacitor does not charge instantly — it fills up gradually, like a bucket under a faucet. At first, the bucket is empty and water rushes in fast. As it fills, the flow slows down. The "time constant" RC tells you how fast it fills: bigger capacitor or bigger resistor = slower filling. After one time constant, it is about 63% full. After five, it is essentially full.

11.6 Common Misconceptions

1. "Current is used up by resistors."

Current is conserved — it is the same entering and leaving any resistor or junction. What is "used up" is voltage (energy per charge). A resistor drops the potential; it does not reduce the rate of charge flow. The water analogy helps: water molecules aren't destroyed as they pass through a narrow pipe, but the pressure drops.

2. "Electrons travel at the speed of light through wires."

Individual electrons drift at ~0.1 mm/s. The signal (electric field) propagates at near light speed, but the charge carriers themselves move very slowly. Turn on a hose — water comes out instantly, but a given molecule takes seconds to travel the hose length.

3. "A dead battery has no voltage."

A "dead" battery still has a voltage close to its rated value when measured with a high-resistance voltmeter (no load). What has degraded is its internal resistance — under load, most of the voltage drops internally, and little reaches the external circuit. A 9 V battery might read 8.5 V unloaded but drop to 2 V when trying to deliver 1 A.

4. "Adding a resistor in parallel always increases total resistance."

Adding any resistor in parallel creates an additional path for current and always decreases the equivalent resistance. \(R_{\text{eq}}\) is always less than the smallest individual resistor in the parallel group.

5. "A capacitor is fully charged after one time constant."

After one time constant, a capacitor reaches only ~63% of full charge. It takes approximately \(5\tau\) to reach >99% — the process is exponential, never actually reaching 100% in finite time (though practically complete after \(5\tau\)).

6. "Ohm's law applies to all circuit elements."

Ohm's law (\(V = IR\)) only describes ohmic materials where resistance is constant. Diodes, transistors, fluorescent lamps, and many other devices have nonlinear \(I\)–\(V\) relationships. The relationship \(R = V/I\) always defines resistance, but that resistance may not be constant.

7. "The path of least resistance takes all the current."

Current divides among all available parallel paths, with more current flowing through lower-resistance branches — but some current flows through every path. If your body touches a high-voltage line, you and the wire form parallel paths to ground, and a lethal current can flow through you even though the wire's resistance is much lower.

Topic Summary

• Capacitance \(C = Q/V\). Parallel-plate: \(C = \kappa\epsilon_0 A/d\). Energy stored: \(U = \frac{1}{2}CV^2\).
• Current \(I = \Delta Q/\Delta t\). Ohm's law: \(V = IR\) (ohmic materials only). Drift velocity is very slow (~0.1 mm/s), but the electric signal propagates near \(c\).
• Electrical power \(P = IV = I^2R = V^2/R\). Resistivity \(\rho\) determines resistance: \(R = \rho L/A\).
• Resistors in series add (\(R_{\text{eq}} = R_1 + R2 + \cdots\)); in parallel reduce (\(1/R{\text{eq}} = 1/R_1 + 1/R_2 + \cdots\)). Capacitors: opposite rules.
• Kirchhoff's rules: junction rule (charge conservation, \(\sum I{\text{in}} = \sum I{\text{out}}\)) and loop rule (energy conservation, \(\sum \Delta V = 0\)). These are universal tools for analyzing any DC circuit.
• RC circuits: exponential charging and discharging with time constant \(\tau = RC\). After \(1\tau\): ~63%; after \(5\tau\): >99% complete.

Essential Equations

Concept Check

1. Why does inserting a dielectric increase capacitance?
2. Two identical light bulbs are connected in series to a battery. A third identical bulb is added in parallel to one of them. What happens to the brightness of each bulb?
3. Why does a capacitor block steady DC current but pass AC?
4. In an RC circuit, is the capacitor fully charged after one time constant? Explain.
5. A 100 W light bulb and a 60 W bulb are connected in series to 120 V. Which glows brighter? Why?
6. A copper wire carries 2.0 A. Estimate the drift velocity of electrons. Why does a light turn on instantly despite this slow speed?
7. A capacitor is charged by a battery, then disconnected. A dielectric is inserted. Does the stored energy increase, decrease, or stay the same? Why?
8. In a multi-loop circuit, you measure a current in a branch opposite to your assumed direction. What does a negative current value mean?

Open Educational References

• OpenStax, College Physics, Chapters 19–21: Electric Potential, Current, Circuits
• OpenStax, University Physics, Volume 2, Chapters 8–10: Capacitance, Current, DC Circuits