Unit 3 Electric Circuits: Energy, Power, and Capacitors

Power in Circuits

What “power” means in an electric circuit

Power is the rate at which energy is transferred or transformed. In circuits, electric forces move charge through a potential difference, and that electrical energy gets converted into other forms—thermal energy in resistors, light in bulbs/LEDs, chemical energy while charging a battery, mechanical energy in a motor, etc.

Thinking in terms of energy helps you avoid a common trap: voltage and current by themselves don’t tell you “how fast” energy is being used. For example, a device can have a large voltage across it but almost no current (so it uses almost no power), like a capacitor in steady-state DC.

In circuits, the key relationship is:

P = \frac{\Delta E}{\Delta t}

where P is power (watts, W), \Delta E is energy transferred (joules, J), and \Delta t is time (seconds, s).

Electrical power: connecting voltage, current, and energy transfer

When a charge q moves through a potential difference \Delta V, the electric potential energy change is:

\Delta U = q\Delta V

If charge flows steadily, current is the rate of charge flow:

I = \frac{\Delta q}{\Delta t}

Combine these ideas: energy transferred per unit time equals voltage times current.

P = IV

• P is the power associated with a circuit element
• I is the current through that element
• V is the potential difference across that element

This is one of the most tested relationships because it links measurable circuit quantities directly to energy transfer.

Passive sign convention: when is power positive or negative?

You’ll often hear statements like “the battery supplies power” and “the resistor dissipates power.” That language comes from a sign convention.

• If conventional current enters the higher-potential side of an element (a “voltage drop” in the direction of current), the element absorbs power: it converts electrical energy into some other form (often thermal). In that case, P = IV is positive.
• If current enters the lower-potential side (a “voltage rise” in the direction of current), the element delivers power to the circuit (like an ideal battery does). In that case, the element’s power is negative under the passive convention.

On AP Physics 2, you’re usually not required to do elaborate sign-convention bookkeeping, but you should be able to interpret statements like “power delivered by the battery equals total power dissipated by resistors” for a steady-state circuit.

Power in resistors: three equivalent forms

For a resistor, Ohm’s law relates the voltage across it and current through it:

V = IR

Substitute into P = IV to get two very useful alternatives:

P = I^2R

P = \frac{V^2}{R}

When to use each form:

• P = IV when you know the element’s voltage and current directly.
• P = I^2R when current through a resistor is easiest to find.
• P = \frac{V^2}{R} when voltage across a resistor is easiest to find.

A subtle but crucial idea: for a fixed resistor, power increases with the square of current or voltage. Doubling current makes power four times larger—this is why small increases in current can cause a lot of heating.

Notation and relationship table (power)

Energy usage over time

If a device operates at (approximately) constant power P for a time t, the energy converted is:

E = Pt

This shows up in household electricity contexts (kilowatt-hours) and in circuit problems where a resistor heats something over a time interval.

How power distributes in series and parallel circuits

Power questions often require you to combine circuit analysis (equivalent resistance, current division, voltage division) with a power formula.

Series resistors:

• Same current flows through each resistor.
• Voltage drops divide based on resistance.
• Power in each resistor is P_i = I^2R_i, so the larger resistor dissipates more power because the same I flows through it.

Parallel resistors:

• Same voltage across each branch.
• Currents divide.
• Power in each resistor is P_i = \frac{V^2}{R_i}, so the smaller resistance dissipates more power because it draws more current at the same voltage.

This is a common misconception point: students sometimes think “bigger resistance means more power” in all cases. That’s only true in series (same current). In parallel (same voltage), smaller resistance means more power.

Real-world meaning: ratings, heating, and safety

Resistors, bulbs, and appliances have power ratings (like 60 W or 1200 W). A rating typically means “this device is designed to safely convert roughly this much electrical power into other forms under its intended operating conditions.”

For a resistor, power dissipated usually becomes thermal energy (Joule heating). That heating is not just an afterthought—it’s the physical mechanism behind blown fuses, overheated wires, and why engineers choose wire gauges.

Worked example 1: power in a simple resistor

A 12\ \text{V} battery is connected across a 6\ \Omega resistor.

1) Find the current:

I = \frac{V}{R} = \frac{12}{6} = 2\ \text{A}

2) Find the power dissipated by the resistor (use any form):

P = IV = (2)(12) = 24\ \text{W}

Check with another form:

P = I^2R = (2^2)(6) = 24\ \text{W}

Interpretation: the resistor converts 24\ \text{J} of electrical energy into thermal energy every second.

Worked example 2: comparing power in series vs parallel

Two resistors R_1 = 2\ \Omega and R_2 = 8\ \Omega are connected to a 10\ \text{V} source.

Case A: series

Equivalent resistance:

R_{eq} = 2 + 8 = 10\ \Omega

Current:

I = \frac{10}{10} = 1\ \text{A}

Power in each:

P_1 = I^2R_1 = (1^2)(2) = 2\ \text{W}

P_2 = I^2R_2 = (1^2)(8) = 8\ \text{W}

The larger resistor dissipates more power in series.

Case B: parallel

Voltage across each is 10\ \text{V}.

P_1 = \frac{V^2}{R_1} = \frac{10^2}{2} = 50\ \text{W}

P_2 = \frac{V^2}{R_2} = \frac{10^2}{8} = 12.5\ \text{W}

Now the smaller resistor dissipates more power in parallel.

Exam Focus

• Typical question patterns:
  • Calculate power dissipated by one element (often a resistor) using circuit info from series-parallel reduction.
  • Compare how power changes when a circuit is rewired (series to parallel) or when resistance changes.
  • Use energy ideas (like E = Pt) to connect power to heating or energy consumption over time.
• Common mistakes:
  • Using P = I^2R or P = \frac{V^2}{R} with the wrong “known” quantity (mixing up whether the resistor has the same I or the same V as others).
  • Forgetting that the V in P = IV is the voltage across the element, not necessarily the battery voltage.
  • Treating “higher resistance means higher power” as universally true (it depends on whether I or V is held constant by the circuit arrangement).

Capacitors in Circuits

What a capacitor is (and why circuits need them)

A capacitor is a circuit element that stores separated charge and therefore stores energy in an electric field. Physically, it’s often two conductors separated by an insulator. When connected in a circuit, a capacitor can accumulate charge on its plates, creating a potential difference across it.

Capacitors matter because they let circuits:

• store energy temporarily (camera flashes, defibrillators)
• smooth or filter voltages (power supplies)
• create time-dependent behavior (timers, signal processing)

In AP Physics 2, capacitors are a core example of how circuits can behave very differently over time, especially during charging and discharging.

Capacitance: the defining relationship

Capacitance C tells you how much charge a capacitor stores per volt across it:

C = \frac{Q}{V}

• C in farads (F)
• Q is the magnitude of charge on one plate (coulombs, C)
• V is the potential difference across the capacitor (volts, V)

Rearrangements you will use constantly:

Q = CV

Key interpretation: for a given capacitor, charge and voltage rise and fall together. If the voltage across the capacitor is zero, the net stored charge separation is zero.

Energy stored in a capacitor

Charging a capacitor requires work because you are separating charge against electric forces. That work becomes energy stored in the electric field.

The standard energy expressions are:

U = \frac{1}{2}CV^2

U = \frac{1}{2}QV

U = \frac{Q^2}{2C}

You don’t need to re-derive these on the exam, but you should understand the logic: as you add charge, the voltage increases, so each additional bit of charge requires more work than the previous bit.

A common misconception is to treat a capacitor like a battery (a fixed voltage source). A capacitor’s voltage is generally not fixed; it depends on how much charge it currently stores.

Capacitors in series and parallel

When multiple capacitors are connected, you often replace them with an equivalent capacitance C_{eq} that has the same overall relationship between total charge and total voltage for that connection.

Parallel capacitors

In parallel, capacitors share the same potential difference V. The total charge stored is the sum of charges on each capacitor:

Q_{tot} = Q_1 + Q_2 + \cdots

Using Q = CV with the same V:

C_{eq} = C_1 + C_2 + \cdots

Intuition: adding parallel capacitors increases the effective plate area, so the system can store more charge per volt.

Series capacitors

In series, the same amount of charge ends up on each capacitor (because charge that leaves one plate must appear on the next conductor in the chain). The voltages add:

V_{tot} = V_1 + V_2 + \cdots

With Q common and V_i = \frac{Q}{C_i}:

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Intuition: series increases the effective separation between plates, reducing the ability to store charge per volt.

A frequent mistake is to swap the series/parallel rules with the resistor rules. Resistors and capacitors behave oppositely in series/parallel combinations.

Voltage and charge sharing in series capacitors

In series, all capacitors carry the same Q, but their voltages differ:

V_i = \frac{Q}{C_i}

So the smaller capacitance gets the larger voltage. This matters in real designs because a capacitor has a maximum safe voltage rating; series strings can be used to handle higher voltages, but the voltage may not divide evenly unless capacitors are matched.

How capacitors behave in DC circuits over time

The most important behavioral idea is this:

• A capacitor resists changes in voltage across it.
• In a DC circuit, after a long time (steady state), an ideal capacitor acts like an open circuit (no current through that branch).

Why does current stop? As the capacitor charges, its voltage rises until it matches the circuit conditions that make the net driving force for current through the charging path go to zero.

This time dependence is why AP problems often specify “immediately after the switch is closed” versus “a long time after the switch is closed.” Those are different circuit states.

RC circuits: charging and discharging

An RC circuit contains a resistor and a capacitor, and it exhibits exponential charging/discharging. The key timescale is the time constant:

\tau = RC

• R is the resistance in ohms
• C is the capacitance in farads
• \tau is in seconds

Conceptually, \tau tells you how quickly the capacitor approaches its final voltage.

Charging (capacitor initially uncharged)

For a series RC connected to an ideal battery of emf \mathcal{E}, the capacitor voltage rises toward \mathcal{E}:

V_C(t) = \mathcal{E}\left(1 - e^{-t/RC}\right)

Charge on the capacitor:

Q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right)

Current in the circuit decreases over time:

I(t) = \frac{\mathcal{E}}{R}e^{-t/RC}

Important limiting cases:

• At t = 0, the capacitor behaves like a wire (approximately). Current is maximum: I(0) = \frac{\mathcal{E}}{R}.
• As t \to \infty, current goes to zero and V_C \to \mathcal{E}.

Students often mix up what “starts at zero.” For charging, the capacitor’s voltage starts at 0 (if uncharged), while current starts at its maximum.

Discharging (capacitor initially charged)

If a charged capacitor discharges through a resistor, the voltage and charge decay exponentially:

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

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

Current magnitude also decays:

I(t) = \frac{V_0}{R}e^{-t/RC}

The direction of current during discharge is opposite the charging direction (it is driven by the capacitor’s stored energy).

“One time constant” meaning (a useful benchmark)

Because the behavior is exponential, a time constant is a convenient reference:

• After one time constant t = RC, a charging capacitor reaches about 63% of its final value.
• After one time constant, a discharging capacitor drops to about 37% of its initial value.

You don’t always need those percentages, but they help you sanity-check graphs and proportional reasoning questions.

Power and energy during capacitor charging

Even though a battery might deliver energy during charging, not all of that energy ends up stored in the capacitor. In a simple series RC charging process:

• The capacitor stores final energy U = \frac{1}{2}C\mathcal{E}^2.
• The resistor dissipates the rest as thermal energy over the charging interval.

This surprises many students because they expect “battery energy = capacitor energy.” The resistor matters because current flows through it while charging, and it dissipates power P_R(t) = I(t)^2R at every moment.

Worked example 1: equivalent capacitance and final charges

Two capacitors C_1 = 3\ \mu\text{F} and C_2 = 6\ \mu\text{F} are connected in parallel to a 12\ \text{V} battery.

1) Equivalent capacitance:

C_{eq} = C_1 + C_2 = 9\ \mu\text{F}

2) Charge on each capacitor (same voltage in parallel):

Q_1 = C_1V = (3\ \mu\text{F})(12\ \text{V}) = 36\ \mu\text{C}

Q_2 = C_2V = (6\ \mu\text{F})(12\ \text{V}) = 72\ \mu\text{C}

3) Total charge drawn from the battery:

Q_{tot} = C_{eq}V = (9\ \mu\text{F})(12\ \text{V}) = 108\ \mu\text{C}

Notice Q_{tot} = Q_1 + Q_2, which is a good consistency check.

Worked example 2: series capacitors and voltage division

Two capacitors C_1 = 2\ \mu\text{F} and C_2 = 4\ \mu\text{F} are connected in series across a 12\ \text{V} battery.

1) Equivalent capacitance:

\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}

So:

C_{eq} = \frac{4}{3}\ \mu\text{F}

2) Charge on the series string:

Q = C_{eq}V = \left(\frac{4}{3}\ \mu\text{F}\right)(12\ \text{V}) = 16\ \mu\text{C}

In series, each capacitor has the same magnitude of charge Q = 16\ \mu\text{C}.

3) Voltage across each:

V_1 = \frac{Q}{C_1} = \frac{16\ \mu\text{C}}{2\ \mu\text{F}} = 8\ \text{V}

V_2 = \frac{Q}{C_2} = \frac{16\ \mu\text{C}}{4\ \mu\text{F}} = 4\ \text{V}

Check: V_1 + V_2 = 12\ \text{V}.

This illustrates the key idea: smaller capacitance gets larger voltage in series.

Worked example 3: RC time constant and capacitor voltage vs time

A 10\ \text{k}\Omega resistor and a 50\ \mu\text{F} capacitor are connected in series to a 9\ \text{V} battery. The capacitor starts uncharged.

1) Time constant:

\tau = RC = (10\times 10^3\ \Omega)(50\times 10^{-6}\ \text{F}) = 0.50\ \text{s}

2) Capacitor voltage after t = 1.0\ \text{s}:

V_C(t) = 9\left(1 - e^{-t/RC}\right) = 9\left(1 - e^{-1.0/0.50}\right) = 9\left(1 - e^{-2}\right)

Numerically, e^{-2} \approx 0.135, so:

V_C \approx 9(1 - 0.135) \approx 7.8\ \text{V}

Interpretation: after 2 time constants, the capacitor is mostly charged but not fully at 9 V.

Graph reasoning: what AP questions often really test

AP questions frequently test whether you understand qualitative shapes:

• During charging, I(t) decreases exponentially from its maximum value to 0.
• During charging, V_C(t) rises asymptotically toward \mathcal{E}.
• During discharging, V_C(t) decays exponentially toward 0.

A classic mistake is drawing linear graphs (straight lines) instead of exponential curves, especially for current.

Exam Focus

• Typical question patterns:
  • Find C_{eq} for series/parallel capacitor networks, then compute charges and voltages on individual capacitors.
  • RC transient questions that ask for \tau = RC, the current or voltage immediately after switching, and the long-time (steady-state) behavior.
  • Interpret or match graphs of V_C(t) and I(t) for charging/discharging.
• Common mistakes:
  • Swapping series/parallel rules with resistor rules (remember: capacitors add in parallel, reciprocals add in series).
  • Assuming a capacitor in a DC circuit always has current through it; in steady state, an ideal capacitor carries zero current.
  • Confusing what is the same in series vs parallel: in series capacitors, Q is the same; in parallel capacitors, V is the same.