Unit 3: Electric Circuits

Electric current, potential difference, resistance, and other circuit quantities

A circuit is a closed pathway that allows electric charges to move under the influence of an electric field. In electrostatics, charges are often stationary and you track the electric potential they create. In circuits, the key shift is that charges are in motion, and that motion is organized by conductors, sources (like batteries), and components that control how much charge flows.

Circuit quantities (what they mean)

Voltage is the difference in electric potential between two points in a circuit. It is measured in volts (V) and represents the “push” (energy per unit charge) that can drive current.

Current is the rate of flow of electric charge through a cross-section of a conductor. It is measured in amperes (A).

Resistance is the opposition to the flow of electric current. It is measured in ohms.

Power is the rate at which energy is transferred or converted in a circuit. It is measured in watts (W).

Frequency is the number of cycles per second in an alternating current (AC) signal. It is measured in hertz (Hz).

Impedance is the total opposition to current in AC circuits, incorporating not just resistance but also effects from capacitance and inductance. It is measured in ohms and commonly represented by the symbol Z.

Electric current: what it is and what it is not

Electric current is defined as the rate at which charge passes a cross-section:

I = \frac{dQ}{dt}

Two ideas that commonly trip students up:

1. Current is not the same thing as charge. Charge is “how much,” current is “how fast.”
2. The direction of current is conventional. By definition, conventional current points in the direction positive charge would move. In metal wires, the actual moving charges are electrons (negative), so electron drift is opposite the direction of conventional current. Circuit equations use conventional current and still work perfectly.

Current can be direct current (DC), where charge flows in one direction only, or alternating current (AC), where the direction changes periodically.

Current at the microscopic level (why resistors heat up)

In a metal, many electrons are free to move. When you apply an electric field in the wire, electrons accelerate, but they constantly collide with the metal lattice (ions) and impurities. Those collisions convert organized electrical energy into random thermal energy.

This microscopic picture explains two macroscopic facts: a larger electric field tends to produce a larger current, and collisions cause energy dissipation, which we model as resistance. A helpful misconception-buster is that electrons do not “race” around the circuit carrying energy like projectiles; their drift speed is typically small. Energy transfer is mediated by electric fields established around the circuit.

Electric potential difference (voltage)

Electric potential difference between two points is the change in electric potential energy per unit charge:

\Delta V = \frac{\Delta U}{q}

In circuits, it is essential to describe voltage between two points (or across an element) and to keep a consistent sign convention when labeling terminal potentials.

Resistance and Ohm’s law (element model)

For an ideal (ohmic) resistor, the circuit-level relationship is:

V = IR

Ohm’s law is not universal for all components. It applies well to many resistors over a wide range, but not to non-linear devices such as diodes, filament bulbs (temperature-dependent), and some sensors.

A useful materials-and-geometry model for a uniform wire is:

R = \rho \frac{L}{A}

Resistance depends on material (resistivity), length, cross-sectional area, and temperature. In many conductors, higher temperature increases resistance, which (for a fixed applied voltage) decreases current.

Types of resistors and why they matter

Resistors appear in multiple forms depending on function:

• Fixed resistors have a set resistance value.
• Variable resistors can be adjusted.
• Thermistors have resistance that varies with temperature.
• Light-dependent resistors (LDRs) have resistance that varies with light intensity.

Resistors are used to control current and voltage in circuits, in sensors (temperature, light, etc.), in heating elements (converting electrical energy into heat), and in motor circuits to help control speed and torque.

Power and energy in circuits

Electric power is the rate of energy transfer:

P = IV

The watt is:

1\,\text{W} = 1\,\text{J/s}

Common larger units include kilowatts (kW) and megawatts (MW).

Using Ohm’s law for a resistor gives equivalent power forms:

P = I^2R

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

These forms are powerful conceptually, but you must match them to what is actually held constant. For example, at constant voltage, increasing resistance decreases power because of P = V^2/R. At constant current, increasing resistance increases power because of P = I^2R.

Energy delivered or dissipated over time t is:

E = Pt

A common conceptual mistake is thinking a battery “supplies current.” A battery supplies energy per unit charge (emf). The actual current depends on the entire circuit.

Applications of current (why circuits show up everywhere)

Current is central to electric power generation, electronics (computers, televisions, smartphones), transportation (electric vehicles and trains), and medical devices (pacemakers and defibrillators).

Worked example: interpreting power correctly

A 6.0\,\Omega resistor has a potential difference of 12\,\text{V} across it.

1) Find the current:

I = \frac{V}{R} = \frac{12}{6.0} = 2.0\,\text{A}

2) Find the power dissipated:

P = IV = (2.0)(12) = 24\,\text{W}

Self-check with a different form:

P = \frac{V^2}{R} = \frac{144}{6.0} = 24\,\text{W}

This same computation is often presented as “if a circuit has 12 V and 2 A, then the power is 24 W,” using P = VI.

Exam Focus

• Typical question patterns:
  • Given V and R (or geometry and \rho), find I and power dissipated.
  • Compare brightness or heating when circuit configuration changes (power changes).
  • Conceptual questions distinguishing charge, current, voltage, and energy.
  • Quick power calculations using P = VI and unit reasoning (watts as joules per second).
• Common mistakes:
  • Treating Ohm’s law as valid for any device without being told it is ohmic.
  • Mixing up “voltage at a point” with “voltage across an element” (voltage is always between two points).
  • Using a power form without checking what is held constant (constant V vs constant I situations).
  • Forgetting that temperature can change resistance, changing current for a fixed applied voltage.

Circuit symbols and circuit measuring tools

Circuit diagrams rely on standardized symbols so you can interpret topology quickly. Even when a diagram is stylized, the core skill is identifying which elements share nodes (same potential in ideal-wire assumptions) and which elements are truly in series or in parallel.

Common circuit symbols (conceptual checklist)

You are expected to recognize the idea of symbolic representations for common components such as resistors, capacitors, batteries (sources), switches, and measuring devices (ammeters and voltmeters). The key payoff is not artistry; it is reading node connections correctly.

Circuit measuring tools (what they measure)

Circuit measuring tools are used to measure electrical parameters for troubleshooting, testing, and design.

• Multimeter: measures voltage, current, and resistance; used for troubleshooting and continuity checks; comes in analog and digital versions.
• Oscilloscope: displays voltage signals over time; used to analyze waveforms; comes in analog and digital versions.
• Function generator: produces test waveforms such as sine, square, and triangle waves.
• Logic analyzer: captures and analyzes digital signals.
• Power supply: provides a controlled constant voltage or constant current for testing.
• LCR meter: measures inductance, capacitance, and resistance.

Exam Focus

• Typical question patterns:
  • Interpret which components share the same two nodes (parallel) by reading a schematic.
  • Identify what measurement device you would use for a given quantity (time-varying signal, capacitance, etc.).
• Common mistakes:
  • Misreading a diagram because of its visual layout instead of tracing nodes and connections.
  • Assuming a multimeter set to measure current can be used like a voltmeter (meter mode matters in real labs).

EMF, batteries, and internal resistance

To sustain a steady current in a closed loop, you need a component that raises electric potential energy. Otherwise charges would settle into electrostatic equilibrium and stop moving. That role is played by sources such as batteries and ideal power supplies.

EMF: energy per unit charge supplied by a source

Electromotive force, written \mathcal{E}, is the energy per unit charge a source provides to move charge from lower potential to higher potential inside the source (against the electrostatic force).

\mathcal{E} = \frac{W}{q}

Even though the name includes “force,” emf is not a force; it is measured in volts because it is energy per charge.

The emf of a battery is the maximum voltage it can provide when no current is flowing through it (open circuit). In real batteries, emf can be affected by factors such as temperature, electrolyte concentration, and the electrode materials.

Ideal vs real sources and internal resistance

Real batteries are non-ideal and can be modeled as an ideal emf source in series with an internal resistance r. When a battery delivers current I to an external circuit, the terminal voltage is reduced by the internal drop:

V_{\text{terminal}} = \mathcal{E} - Ir

The internal voltage drop itself follows the same resistor rule:

V_{\text{internal drop}} = Ir

Two extreme cases help build intuition:

• Open circuit (no current, I = 0):

V_{\text{terminal}} = \mathcal{E}

This is why the emf can be measured using a voltmeter across the battery terminals when no current is flowing.

• Short circuit (external resistance near zero):

I \approx \frac{\mathcal{E}}{r}

This is why shorting a battery is dangerous: large current causes large power dissipation inside the battery.

If the battery is being charged (current forced into the battery opposite the usual direction), the terminal voltage can exceed \mathcal{E}; AP-style problems will typically state the situation clearly.

Measuring current and voltage (ideal meters)

• An ammeter measures current and is placed in series with the element whose current you want. The ideal ammeter has negligible resistance.
• A voltmeter measures potential difference and is placed in parallel with the element whose voltage you want. The ideal voltmeter has extremely large resistance.

Placing an ammeter in parallel can effectively short a circuit; placing a voltmeter in series can block current.

Worked example: internal resistance and terminal voltage

A battery has \mathcal{E} = 12\,\text{V} and internal resistance r = 1.0\,\Omega. It is connected to an external resistor R = 5.0\,\Omega.

1) Current (series resistances add):

I = \frac{\mathcal{E}}{R + r} = \frac{12}{5.0 + 1.0} = 2.0\,\text{A}

2) Terminal voltage:

V_{\text{terminal}} = \mathcal{E} - Ir = 12 - (2.0)(1.0) = 10\,\text{V}

3) Physical interpretation: the source provides 12 V of potential rise internally, but 2 V is lost across internal resistance, leaving 10 V for the external circuit.

Exam Focus

• Typical question patterns:
  • Determine current and terminal voltage for a battery with internal resistance.
  • Compute short-circuit current or explain why it is large.
  • Conceptual: distinguish \mathcal{E} from measured terminal voltage.
  • Recognize open-circuit measurement as a way to read emf with a voltmeter.
• Common mistakes:
  • Treating emf and terminal voltage as always equal even when current flows.
  • Forgetting internal resistance is in series with the external circuit.
  • Incorrect meter placement (ammeter parallel, voltmeter series).

Combining resistors and capacitors (equivalent circuits)

Much of circuit problem-solving is reducing a network into simpler equivalent elements. This works because of two physical constraints:

• Components in series share the same current because there is only one path for charge flow.
• Components in parallel share the same voltage because they connect to the same two nodes.

Series and parallel resistors

A series circuit is one loop: current flows through each component in turn, so each element carries the same current. The total resistance is the sum of the resistances:

R_{\text{eq}} = R_1 + R_2 + \cdots

In a series chain, the voltage across each resistor is proportional to its resistance because the current is the same through each and V = IR.

A parallel circuit has separate branches: the voltage across each branch is the same, and current divides among branches. The equivalent resistance satisfies:

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

In parallel, the total resistance is less than the smallest branch resistance. Current through each branch is larger for smaller resistance; equivalently, current is proportional to conductance (inverse resistance).

Capacitors: charge storage and circuit role

A capacitor stores energy in an electric field between conductors separated by an insulator. In circuit models:

Q = C\Delta V

A capacitor does not “hold current” permanently; current is what changes the stored charge:

I = \frac{dQ}{dt}

For constant capacitance this also implies:

I = C\frac{dV}{dt}

Series and parallel capacitors

Capacitors combine opposite to resistors.

Parallel capacitors share the same voltage, and their charges add:

C_{\text{eq}} = C_1 + C_2 + \cdots

Series capacitors carry the same charge magnitude, and their voltages add:

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

For series capacitors, since V_i = Q/C_i, the voltage across each capacitor is inversely proportional to its capacitance. The smallest capacitance in the series string has the largest voltage across it.

Energy stored in a capacitor

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

Equivalent forms:

U = \frac{1}{2}QV

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

A capacitor does not create energy; energy is supplied by the source during charging and stored in the electric field.

Worked example: mixed resistor network

A resistor R_1 = 2.0\,\Omega is in series with a parallel combination of R_2 = 6.0\,\Omega and R_3 = 3.0\,\Omega.

1) Equivalent resistance of the parallel part:

\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0}

R_p = 2.0\,\Omega

Then:

R_{\text{eq}} = R_1 + R_p = 2.0 + 2.0 = 4.0\,\Omega

2) If connected to an ideal 12\,\text{V} source:

I_{\text{total}} = \frac{12}{4.0} = 3.0\,\text{A}

3) Current through each parallel branch:

V_1 = I_{\text{total}}R_1 = (3.0)(2.0) = 6.0\,\text{V}

So the parallel network has:

V_p = 12 - 6.0 = 6.0\,\text{V}

Branch currents:

I_2 = \frac{V_p}{R_2} = \frac{6.0}{6.0} = 1.0\,\text{A}

I_3 = \frac{V_p}{R_3} = \frac{6.0}{3.0} = 2.0\,\text{A}

Check:

I_2 + I_3 = 3.0\,\text{A}

Worked example: capacitors in series with a voltage source

Two capacitors C_1 = 2.0\,\mu\text{F} and C_2 = 6.0\,\mu\text{F} are connected in series across 12\,\text{V}.

1) Equivalent capacitance:

\frac{1}{C_{\text{eq}}} = \frac{1}{2.0} + \frac{1}{6.0} = \frac{4}{6.0}

C_{\text{eq}} = 1.5\,\mu\text{F}

2) Series charge magnitude:

Q = C_{\text{eq}}V = (1.5\times 10^{-6})(12) = 1.8\times 10^{-5}\,\text{C}

3) Individual voltages:

V_1 = \frac{Q}{C_1} = \frac{1.8\times 10^{-5}}{2.0\times 10^{-6}} = 9.0\,\text{V}

V_2 = \frac{Q}{C_2} = \frac{1.8\times 10^{-5}}{6.0\times 10^{-6}} = 3.0\,\text{V}

Check:

V_1 + V_2 = 12\,\text{V}

Exam Focus

• Typical question patterns:
  • Reduce circuits using series-parallel equivalents, then find currents and voltages.
  • Determine how adding a parallel branch changes total current and power.
  • Use proportional reasoning: series resistors share current, parallel resistors share voltage.
  • Capacitor networks: find C_{\text{eq}}, then find series charge and individual capacitor voltages.
• Common mistakes:
  • Mixing series and parallel rules (especially for capacitors, which combine opposite to resistors).
  • Assuming current splits equally in parallel; it splits based on resistance.
  • Forgetting series capacitors carry the same charge magnitude, not the same voltage.
  • Forgetting parallel resistors must give an equivalent smaller than the smallest branch resistance.

Kirchhoff’s rules (loop and junction analysis)

Series-parallel reduction is fast, but it only works when the circuit can actually be simplified that way. Many circuits require Kirchhoff’s rules, which come directly from conservation laws.

• The junction rule comes from conservation of charge.
• The loop rule comes from conservation of energy.

Kirchhoff’s Voltage Law (KVL) assumes there are no changing magnetic fields threading the loop. If magnetic flux through a loop changes, induced emf effects (Faraday’s law of induction) can add non-conservative potential changes that require more advanced analysis.

Nodes, branches, and loops (how to read circuit topology)

A node is a point where two or more elements connect and share the same electric potential (ideal wire assumption). A branch is a path between nodes containing elements. A loop is any closed path.

Kirchhoff’s junction rule (KCL)

At any junction, the algebraic sum of currents is zero:

\sum I = 0

This is often written as “current in equals current out.” You choose current directions; a negative result means the actual direction is opposite your assumption.

Kirchhoff’s loop rule (KVL)

Around any closed loop, the algebraic sum of potential changes is zero:

\sum \Delta V = 0

Sign conventions that keep you consistent:

• Crossing an ideal source from negative to positive terminal contributes +\mathcal{E}.
• Crossing from positive to negative contributes -\mathcal{E}.
• For a resistor, traversing in the same direction as the assumed current gives a drop -IR.
• Traversing opposite the assumed current gives a rise +IR.

Strategy for solving Kirchhoff problems

A reliable procedure:

1. Label currents in each branch (choose directions consistently).
2. Write junction equations at key nodes.
3. Write independent loop equations (avoid redundant loops).
4. Solve the resulting system. Negative currents are informative, not “wrong.”

Example of KVL (symbolic loop bookkeeping)

For a single loop containing a source and three resistors with drops across R_1, R_2, and R_3, KVL can be expressed as:

V_1 - V_{R_1} - V_{R_2} - V_{R_3} = 0

This is the same idea as writing “rise from the source equals total drops across resistors” when traversing a loop.

Worked example: two-loop circuit with shared resistor

A circuit has two loops sharing a resistor. Left loop: battery \mathcal{E}_1 = 12\,\text{V} in series with R_1 = 4.0\,\Omega and a shared resistor R_3 = 2.0\,\Omega. Right loop: battery \mathcal{E}_2 = 6.0\,\text{V} in series with R_2 = 3.0\,\Omega and the same shared resistor R_3. Assume both batteries oriented to drive clockwise currents in their respective loops.

Let mesh currents be I_1 clockwise in left loop and I_2 clockwise in right loop. The shared resistor current is I_1 - I_2 (based on these assumed loop directions).

Left loop:

+\mathcal{E}_1 - I_1R_1 - (I_1 - I_2)R_3 = 0

Substitute values:

12 - 4.0I_1 - 2.0(I_1 - I_2) = 0

Simplify:

12 - 6.0I_1 + 2.0I_2 = 0

Right loop:

+\mathcal{E}_2 - I_2R_2 - (I_2 - I_1)R_3 = 0

Substitute values:

6 - 3.0I_2 - 2.0(I_2 - I_1) = 0

Simplify:

6 + 2.0I_1 - 5.0I_2 = 0

Solve the system:

6.0I_1 - 2.0I_2 = 12

2.0I_1 - 5.0I_2 = -6

Multiply the second equation by 3:

6.0I_1 - 15.0I_2 = -18

Subtract the first equation:

-13.0I_2 = -30

I_2 = 2.31\,\text{A}

Then from 2.0I_1 - 5.0I_2 = -6:

2.0I_1 - 11.55 = -6

I_1 = 2.78\,\text{A}

Shared resistor current (left-loop direction):

I_3 = I_1 - I_2 = 0.47\,\text{A}

Interpretation: both loops drive clockwise, but the stronger left source produces a slightly larger loop current, so net current through the shared resistor follows the left-loop direction.

Common conceptual traps in Kirchhoff problems

• “Current gets used up.” False. Charge is conserved. Energy is transferred.
• Assuming a battery enforces a fixed current. A battery enforces an emf; current follows from the full network.
• Confusing sign conventions. Your assumed directions can be arbitrary; consistency is what matters.

Exam Focus

• Typical question patterns:
  • Solve multi-loop circuits with two sources and shared resistors using Kirchhoff’s rules.
  • Find a particular branch current, then compute power in an element.
  • Conceptual: determine which direction current flows in a branch before doing full algebra.
  • Recognize when KVL assumptions fail conceptually (changing magnetic flux implies induced emf ideas).
• Common mistakes:
  • Sign errors when traversing resistors relative to assumed current direction.
  • Writing too many dependent loop equations and getting stuck in algebra.
  • Forgetting that elements connected by ideal wire share the same node potential.

RC circuits: transients, time constant, and steady-state behavior

Resistor networks are often “static” in idealized DC analysis: currents settle immediately to steady values. Adding a capacitor makes the circuit dynamic because capacitor voltage depends on stored charge, and stored charge changes over time.

What happens physically when a capacitor charges?

In a series resistor-capacitor loop connected to a battery, an initially uncharged capacitor starts with nearly zero voltage across it, so initially most of the battery voltage appears across the resistor and the current is large. As the capacitor charges, its voltage rises, leaving less voltage across the resistor; current decreases and approaches zero. In DC steady state, a capacitor behaves like an open circuit (no current through it), even though it can hold a nonzero voltage.

The time constant and exponential behavior

For a simple series RC circuit, the time constant is:

\tau = RC

Charging from an uncharged capacitor with an ideal source \mathcal{E}:

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

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

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

A key benchmark: at t = \tau, the capacitor voltage has reached about 63.2% of its final value.

Discharging from an initial voltage V_0 with no source:

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

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

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

Another benchmark: at t = \tau during discharge, the capacitor voltage has dropped to about 36.8% of its initial value.

A frequent misconception is thinking the capacitor charges linearly; it does not, because the driving voltage across the resistor shrinks as the capacitor voltage rises.

How to derive the charging equation (setup you should understand)

Kirchhoff’s loop rule for the charging loop gives:

\mathcal{E} - IR - V_C = 0

Using I = dQ/dt and V_C = Q/C:

\mathcal{E} - R\frac{dQ}{dt} - \frac{Q}{C} = 0

Rearrange:

\frac{dQ}{dt} = \frac{\mathcal{E}}{R} - \frac{Q}{RC}

This leads to the exponential charging solution.

Steady-state shortcuts for capacitors (powerful exam tool)

In DC steady state:

• Capacitors carry zero current.
• Any branch containing a capacitor in series is effectively an open branch.
• Capacitors can still have nonzero voltage and store charge.

This lets you find final node voltages using resistor methods, then compute final capacitor charge via Q = CV.

Worked example: charging current and capacitor voltage

A series RC circuit has R = 2.0\times 10^3\,\Omega, C = 5.0\times 10^{-6}\,\text{F}, and \mathcal{E} = 10\,\text{V}. The capacitor starts uncharged.

1) Time constant:

\tau = RC = (2.0\times 10^3)(5.0\times 10^{-6}) = 1.0\times 10^{-2}\,\text{s}

2) Initial current:

I(0) = \frac{\mathcal{E}}{R} = \frac{10}{2.0\times 10^3} = 5.0\times 10^{-3}\,\text{A}

3) Current at t = 0.020\,\text{s}:

I(t) = \frac{\mathcal{E}}{R}e^{-t/(RC)} = 5.0\times 10^{-3}e^{-0.020/0.010}

I(t) = 5.0\times 10^{-3}e^{-2} = 6.8\times 10^{-4}\,\text{A}

4) Capacitor voltage at that time:

V_C(t) = \mathcal{E}\left(1 - e^{-t/(RC)}\right) = 10\left(1 - e^{-2}\right)

V_C(t) = 8.65\,\text{V}

Worked example: steady-state capacitor in a resistor divider

A capacitor is connected in parallel with R_2 in a two-resistor divider: R_1 in series with R_2 across a battery. After a long time (DC steady state), the capacitor draws no current, so it does not change the resistor currents. The capacitor voltage equals the voltage across R_2:

V_{R_2} = \mathcal{E}\frac{R_2}{R_1 + R_2}

Thus:

V_C = \mathcal{E}\frac{R_2}{R_1 + R_2}

Final capacitor charge:

Q_f = CV_C

RC circuits in signal applications: filters and oscillators

RC circuits are used in filters, timing circuits, and oscillators.

• RC filters can be designed as high-pass (pass high frequencies, block low frequencies) or low-pass (pass low frequencies, block high frequencies). The cutoff frequency is the frequency at which the filter begins to significantly attenuate the signal.
• RC oscillators can generate periodic waveforms; the output frequency is determined by the resistor and capacitor values used.

These applications rely on the same core fact: capacitor voltage cannot change instantly without requiring extremely large current.

Exam Focus

• Typical question patterns:
  • Determine \tau and use exponentials to find I(t), V_C(t), or Q(t).
  • Use steady-state reasoning to find final capacitor voltage in a resistive network.
  • Interpret graphs of I(t) and V_C(t) (exponential decay vs rise).
  • Use the 63.2% (charging) and 36.8% (discharging) benchmarks at one time constant.
  • Conceptual questions connecting RC behavior to filtering or timing.
• Common mistakes:
  • Treating the capacitor like a resistor (using V = IR across it).
  • Forgetting the steady-state rule: capacitor current goes to zero for DC.
  • Plugging time into the exponent with the wrong sign or using e^{-t/(RC)} vs 1 - e^{-t/(RC)} in the wrong place.

Multi-capacitor RC behavior and energy considerations

Even when RC circuits look more complex than a single series resistor and capacitor, the same ideas still drive the solution: identify equivalent capacitances and the effective resistance seen by the capacitor, then apply first-order exponential behavior when appropriate.

When can you treat a circuit as “one time constant”?

A circuit behaves like a first-order RC system (single exponential) when it has one effective capacitor energy storage element (or multiple capacitors that can be combined into an equivalent capacitor) and the rest of the network can be replaced by a single equivalent resistance seen by the capacitor.

A practical method is to use a Thevenin-style viewpoint from the capacitor’s terminals:

• Replace the rest of the circuit by an equivalent source voltage V_{\text{th}} and equivalent resistance R_{\text{th}}.
• Then the time constant is:

\tau = R_{\text{th}}C

Operationally, a common technique is to “turn off” ideal voltage sources (replace them with wires) to find the resistance seen from the capacitor terminals.

Energy flow during charging: where does the battery’s energy go?

If a capacitor is charged to voltage V, the stored energy is:

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

But the battery delivers energy:

E_{\text{battery}} = QV = (CV)V = CV^2

The difference is:

E_{\text{battery}} - U_C = \frac{1}{2}CV^2

That “missing half” is dissipated as thermal energy in the resistor during the charging process.

Worked example: discharge energy and power trend

A capacitor C = 10\,\mu\text{F} is initially charged to V_0 = 20\,\text{V} and then discharged through a resistor R = 1.0\times 10^3\,\Omega.

1) Initial stored energy:

U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(10\times 10^{-6})(400) = 2.0\times 10^{-3}\,\text{J}

2) Time constant:

\tau = RC = (1.0\times 10^3)(10\times 10^{-6}) = 1.0\times 10^{-2}\,\text{s}

3) Current magnitude:

I(t) = \frac{V_0}{R}e^{-t/(RC)} = \frac{20}{1.0\times 10^3}e^{-t/0.010}

I(t) = 2.0\times 10^{-2}e^{-t/0.010}\,\text{A}

4) Power dissipated in the resistor:

P(t) = I(t)^2R

So:

P(t) = \left(2.0\times 10^{-2}e^{-t/0.010}\right)^2(1.0\times 10^3)

P(t) = 0.40e^{-2t/0.010}\,\text{W}

Power decays faster than current because it depends on the square of current.

Practical interpretation: why RC circuits matter

RC behavior underlies timing and delay circuits, basic filtering (such as smoothing rapid voltage changes with low-pass behavior), and capacitive sensing, where changes in capacitance change time constants and response.

Exam Focus

• Typical question patterns:
  • Energy accounting in capacitor charging: compare battery energy to stored capacitor energy.
  • Determine effective \tau by finding the equivalent resistance seen by a capacitor.
  • Analyze discharge power or current decay using exponential relationships.
• Common mistakes:
  • Assuming all supplied energy becomes stored capacitor energy (ignoring resistive heating).
  • Using \tau = RC with the wrong resistance (not the effective resistance seen by the capacitor).
  • Forgetting that power depends on I^2, giving a different exponential rate than current.