Circuit Elements

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

Introduction: Thinking About Circuits

A circuit is a closed loop that lets charge flow continuously. A useful (imperfect) analogy: voltage $\leftrightarrow$ pressure, current $\leftrightarrow$ flow rate, resistance $\leftrightarrow$ pipe narrowness. Circuit questions ask how much charge flows, how much voltage drops, or how much energy is stored or dissipated.

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Current: $I = \Delta Q / \Delta t$

Must know

Electric current $I$ is the rate at which charge passes a cross-section of a conductor:

$I = \frac{\Delta Q}{\Delta t}$

where $\Delta Q$ is charge (C) and $\Delta t$ is time (s). The SI unit is the ampere (A), $1 \text{ A} = 1 \text{ C/s}$.

Sign Convention

Must know

Conventional current flows from the positive terminal, through the external circuit, to the negative terminal — the direction positive charge would flow. Electrons move opposite. The MCAT uses conventional current exclusively. (Nerve action potentials are conventional currents carried by ions; $I = \Delta Q/\Delta t$ still applies.) You do not need electron drift velocity.

Quick check: A defibrillator delivers 300 C of charge in 2 s. What current does it produce?

Answer: $I = 300 / 2 = 150 \text{ A}$.

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Electromotive Force and Voltage

Must know

Electric potential $V$ is potential energy per unit charge (volts, V = J/C). What drives current is the potential difference $\Delta V$ (voltage) between two points. Charge flows from high to low potential; the potential difference tells you how much energy each coulomb releases or gains.

Electromotive force ($\mathcal{E}$) is the work per unit charge done by a source (battery, ion pump) to drive charge around the circuit. Despite the name, it is not a force — it is a potential difference in volts.

A real battery has internal resistance $r$ in series with its ideal EMF source. When current $I$ flows, the terminal voltage (what you measure) is:

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

Discharging, $V_{\text{terminal}} < \mathcal{E}$ because some voltage drops across $r$. At open circuit ($I = 0$), $V_{\text{terminal}} = \mathcal{E}$.

Quick check: A battery has $\mathcal{E} = 12$ V and $r = 2\ \Omega$. When it drives $I = 2$ A, what is the terminal voltage?

Answer: $V_{\text{terminal}} = 12 - (2)(2) = 8 \text{ V}$.

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Resistance

Must know

Resistance $R$ measures how much an element opposes current. Unit: ohm ($\Omega = 1 \text{ V/A}$). Higher $R$ means less current for the same voltage.

Resistivity: $R = \rho L/A$

Must know

Resistance depends on geometry and the material's intrinsic resistivity $\rho$ ($\Omega \cdot \text{m}$):

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

Longer wire → higher $R$ (more chances to collide); thicker wire → lower $R$ (more parallel pathways). Rearranged, $\rho = RA/L$.

Factor	Effect on $R$
Longer wire ($\uparrow L$)	$R$ increases
Thicker wire ($\uparrow A$)	$R$ decreases
More resistive material ($\uparrow \rho$)	$R$ increases
Higher temperature (metals)	$R$ increases

Conductivity $\sigma$ is the inverse of resistivity: $\sigma = 1/\rho$ (units S/m). A good conductor has low $\rho$, high $\sigma$.

Quick check: A wire is replaced by one of the same material but twice the length and half the area. By what factor does resistance change?

Answer: $R' = \rho(2L)/(A/2) = 4\rho L/A = 4R$. Resistance quadruples.

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Ohm's Law

Must know

$V = IR \quad \Longleftrightarrow \quad I = \frac{V}{R}$

Ohm's Law applies to ohmic resistors, for which $R$ is constant. Non-ohmic devices (diodes, transistors) have $R$ that varies with voltage.

Power Dissipation

Must know

Current through a resistor converts electrical energy to heat:

$P = IV = I^2 R = \frac{V^2}{R}$

All three forms are equivalent (sub in Ohm's Law). Units: watts (W = J/s). The MCAT frequently asks which form to use given the available knowns.

Kirchhoff's Laws

Know the logic

Apply these through series/parallel reduction, not multi-loop equations:

• Junction rule (KCL): currents in = currents out (conservation of charge) — why current is the same through series resistors and splits in parallel.
• Loop rule (KVL): potential differences around a closed loop sum to zero (conservation of energy) — why voltage drops add in series.

Quick check: A 9 V battery is connected to a 3 $\Omega$ resistor. Find current and power dissipated.

Answer: $I = 9/3 = 3\ \text{A}$. $P = I^2 R = (9)(3) = 27\ \text{W}$ (or $V^2/R = 81/3 = 27\ \text{W}$). ✓

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Resistors in Series

Must know

Resistors end-to-end in one pathway carry the same current $I$ (one path); voltage drops add:

$V_{\text{total}} = V_1 + V_2 + \cdots \quad \Rightarrow \quad R_{\text{series}} = R_1 + R_2 + \cdots$

Key intuition: adding series resistors always increases total resistance.

Voltage Divider Rule

Know the logic

For two series resistors, voltage divides in proportion to resistance:

$V_1 = V_{\text{total}} \cdot \frac{R_1}{R_1 + R_2}$

Quick check: $R_1 = 2\ \Omega$, $R_2 = 3\ \Omega$, $R_3 = 5\ \Omega$ in series across 20 V. Voltage across $R_2$?

Answer: $R_{\text{total}} = 10\ \Omega$, $I = 20/10 = 2\ \text{A}$, $V_2 = IR_2 = (2)(3) = 6\ \text{V}$. ✓

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Resistors in Parallel

Must know

Resistors sharing the same two endpoints have the same voltage; currents add at the junction:

$I_{\text{total}} = I_1 + I_2 + \cdots \quad \Rightarrow \quad \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

Key intuition: adding parallel resistors always decreases total resistance (more lanes); $R_{\text{eq}}$ is smaller than the smallest resistor. For two resistors, use product over sum:

$R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$

Current Divider Rule

Know the logic

Current divides inversely with resistance (more current takes the easier path):

$I_1 = I_{\text{total}} \cdot \frac{R_2}{R_1 + R_2}$

Note $I_1$ uses $R_2$ (the other resistor) in the numerator — derive from $I_1 R_1 = I_2 R_2$.

Worked Example (Series vs. Parallel): $R_1 = 4\ \Omega$, $R_2 = 12\ \Omega$, 12 V source.

Series: $R_{\text{total}} = 16\ \Omega$; $I = 0.75\ \text{A}$; $P_1 = I^2 R_1 = (0.75)^2(4) = 2.25\ \text{W}$.

Parallel: $R_{\text{total}} = (4)(12)/16 = 3\ \Omega$; $I_{\text{total}} = 4\ \text{A}$; $R_1$ sees the full 12 V, so $P_1 = V^2/R_1 = 144/4 = 36\ \text{W}$.

Sanity check: parallel gives more power because each resistor sees the full source voltage (why household appliances are wired in parallel). ✓

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Capacitance

Must know

A capacitor stores energy: two conductors separated by an insulator. Connected to a source, equal and opposite charge builds on the plates. Capacitance $C$:

$C = \frac{Q}{V}$

Units: farads (F = C/V). A farad is enormous; real capacitors are µF ($10^{-6}$) or pF ($10^{-12}$).

Parallel Plate Capacitor

Must know

Two plates of area $A$ separated by distance $d$, with permittivity $\varepsilon$ between them (vacuum/air: $\varepsilon_0 = 8.85 \times 10^{-12}$):

$C = \frac{\varepsilon_0 A}{d} \quad \text{(vacuum/air)}$

Larger plates and smaller separation → larger $C$. The field between plates is uniform:

$E = \frac{V}{d} = \frac{\sigma}{\varepsilon_0}$

where $\sigma = Q/A$ is surface charge density.

Quick check: If $d$ is doubled and $A$ is halved, by what factor does $C$ change?

Answer: $C \propto A/d$, so $C' = \frac{A/2}{2d} = C/4$. Decreases by a factor of 4.

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Energy Stored in a Charged Capacitor

Must know

$U_C = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}$

All three are equivalent (sub $Q = CV$). The factor of $\frac{1}{2}$ appears because charge builds up gradually — average work is half the final voltage times total charge.

Worked Example: A $10\ \mu\text{F}$ capacitor charged to $6\ \text{V}$:

$U_C = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(10 \times 10^{-6})(36) = 180\ \mu\text{J}$

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Capacitor Behavior in a DC Circuit

Know the logic

You do NOT need the exponential charging/discharging math. Only the two extremes matter:

• At $t = 0$ (uncharged): no voltage across it yet, so it acts like a plain wire (short). Current is maximal, limited by the resistors.
• At steady state (fully charged): no more charge flows, so it acts like an open circuit. That branch carries zero steady-state current and the full source voltage appears across the capacitor.

Quick check: A resistor is in series with a fully charged capacitor in a DC circuit. Current through that branch?

Answer: Zero — a fully charged capacitor blocks DC (open circuit).

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Capacitors in Series and Parallel

Must know

The rules are the opposite of resistors — a notorious MCAT trap.

Series: same charge $Q$ on each (charge can't cross the gap); voltages add:

$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \quad \text{(total } C \text{ decreases)}$

Parallel: same voltage; charges add:

$C_{\text{parallel}} = C_1 + C_2 + \cdots \quad \text{(total } C \text{ increases)}$

Configuration	Resistors	Capacitors
Series	$R_{\text{tot}} = R_1 + R_2 + \cdots$ (increases)	$\frac{1}{C_{\text{tot}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$ (decreases)
Parallel	$\frac{1}{R_{\text{tot}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$ (decreases)	$C_{\text{tot}} = C_1 + C_2 + \cdots$ (increases)

Quick check: Two $6\ \mu\text{F}$ capacitors in series. Total capacitance?

Answer: $\frac{1}{C} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$, so $C = 3\ \mu\text{F}$.

Worked Example: Q and V on Individual Capacitors

Must know

A $2\ \mu\text{F}$ and a $4\ \mu\text{F}$ capacitor in series across 12 V. Each holds the same charge $Q$.

$C_{\text{eq}} = \frac{(2)(4)}{6} = 1.33\ \mu\text{F}, \qquad Q = C_{\text{eq}}V = 16\ \mu\text{C on each}$

$V_{2\mu F} = \frac{16}{2} = 8\ \text{V}, \qquad V_{4\mu F} = \frac{16}{4} = 4\ \text{V}$

Sanity check: voltages sum to 12 V ✓; the smaller capacitor takes the larger voltage share (in series, $V \propto 1/C$). In parallel instead, both see 12 V and charges differ: $24\ \mu\text{C}$ and $48\ \mu\text{C}$.

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Dielectrics

Must know

A dielectric is an insulator inserted between the plates. Its polar molecules align with the field, creating an opposing internal field, so the net field (and voltage, at fixed $Q$) drops and capacitance rises. The dielectric constant $\kappa$ ($\geq 1$, dimensionless) gives the factor:

$C_{\text{dielectric}} = \kappa C_0 = \kappa \frac{\varepsilon_0 A}{d}$

For air $\kappa \approx 1$; water $\kappa \approx 80$ (large dipole; relevant for membranes).

Know the logic

1. Disconnected (Q fixed): $C\uparrow$, so $V = Q/C$ decreases and $U = Q^2/(2C)$ decreases.
2. Connected to battery (V fixed): $C\uparrow$, so $Q = CV$ increases and $U = \frac{1}{2}CV^2$ increases (battery supplies extra charge).

Quick check: $C_0 = 4\ \mu\text{F}$ connected to 10 V; insert $\kappa = 3$ dielectric. New capacitance and energy?

Answer: $C = \kappa C_0 = 12\ \mu\text{F}$. $U = \frac{1}{2}(12 \times 10^{-6})(100) = 600\ \mu\text{J}$.

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Conductivity: Metallic vs. Electrolytic

Must know

Metallic (copper, aluminum): charge carried by free electrons; very low $\rho$. As temperature rises, lattice vibrations scatter electrons more, so $R$ increases. No chemical change.

Electrolytic (aqueous ionic solutions, biological fluids): charge carried by ions (both cations and anions, moving oppositely but both contributing to conventional current). As temperature rises, ion mobility increases, so $R$ decreases. Conductivity rises with ion concentration; reactions can occur at electrodes. This underlies nerve, cardiac, and muscle signaling.

Quick check: Why does saltwater's resistance fall when heated, while copper's rises?

Answer: In copper, heating increases lattice vibrations that scatter electrons ($R\uparrow$). In saltwater, heating raises ion mobility, so ions move faster ($R\downarrow$).

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Ammeters and Voltmeters in Circuits

Must know

Ammeter — measures current; placed in series; ideal resistance zero (so it doesn't add a voltage drop).

Voltmeter — measures voltage; placed in parallel; ideal resistance infinite (so it doesn't divert current).

Instrument	Measures	Placement	Ideal Resistance
Ammeter	Current	Series	$R = 0\ \Omega$
Voltmeter	Voltage	Parallel	$R = \infty\ \Omega$

The point: an ideal meter minimally disturbs the circuit it measures.

Quick check: A student connects a voltmeter in series with a resistor instead of in parallel. What happens to the measured current?

Answer: The voltmeter's huge resistance, placed in series, drops current to nearly zero — it acts like an open switch.

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Common Confusions & Tricks

1. Capacitor vs. resistor combination rules are reversed. Resistors in series add directly; capacitors in series use reciprocals. Resistors in parallel use reciprocals; capacitors in parallel add. Re-derive if unsure: series caps share $Q$ (V adds); parallel caps share $V$ (Q adds).

2. More resistors in parallel = lower resistance, but more capacitors in parallel = higher capacitance. Same math form, opposite physical effect.

3. $R = \rho L/A$ — don't invert the geometry. Thicker and shorter → less $R$. Common error: writing $R \propto A$ when $R \propto 1/A$.

4. EMF vs. terminal voltage. EMF is the "advertised" value; terminal voltage is what's delivered under load. Equal only at $I = 0$; otherwise $V_{\text{terminal}} < \mathcal{E}$.

5. Power formulas — match the knowns. $P = IV$ (know $I$ and $V$); $P = I^2 R$ (series, same $I$); $P = V^2/R$ (parallel, same $V$).

6. Ammeter/voltmeter placement. Swap them and you either short the circuit (ammeter in parallel) or kill the current (voltmeter in series).

7. Dielectric effect depends on whether the battery stays connected. Charge fixed: $V\downarrow$, $U\downarrow$. Voltage fixed: $Q\uparrow$, $U\uparrow$.

8. Temperature and conductivity. Metals: $T\uparrow \Rightarrow R\uparrow$. Electrolytes/semiconductors: $T\uparrow \Rightarrow R\downarrow$.

9. In any parallel circuit, every branch has the same voltage. Use $P = V^2/R$ per branch; smallest $R$ dissipates the most power (why a short trips a breaker).

10. Uniform field in a parallel plate capacitor. $E = V/d$, directed from positive to negative plate; doubling $d$ halves $E$ (at fixed $V$).

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Key Equations

Equation	Variables & When to Use
$I = \Delta Q / \Delta t$	$I$ = current (A), $Q$ = charge (C), $t$ = time (s). Definition of current.
$V_{\text{terminal}} = \mathcal{E} - Ir$	$\mathcal{E}$ = EMF (V), $r$ = internal resistance ($\Omega$). Real battery under load.
$R = \dfrac{\rho L}{A}$	$\rho$ = resistivity ($\Omega\cdot$m), $L$ = length (m), $A$ = area (m²). Geometry of resistance.
$\sigma = 1/\rho$	$\sigma$ = conductivity (S/m). Inverse of resistivity.
$V = IR$	Ohm's Law. $V$ (V), $I$ (A), $R$ ($\Omega$).
$P = IV = I^2R = V^2/R$	$P$ = power (W). Use $I^2R$ in series; $V^2/R$ in parallel.
$R_{\text{series}} = R_1 + R_2 + \cdots$	Resistors in series.
$\dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots$	Resistors in parallel. Two: $R = R_1R_2/(R_1+R_2)$.
$C = Q/V$	Capacitance. $C$ (F), $Q$ (C), $V$ (V).
$C_0 = \dfrac{\varepsilon_0 A}{d}$	Parallel plate, air/vacuum. $A$ (m²), $d$ (m).
$C_{\text{dielectric}} = \kappa C_0$	$\kappa$ = dielectric constant ($\geq 1$).
$U_C = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} = \dfrac{QV}{2}$	Energy stored (J). Choose the form matching your knowns.
$C_{\text{parallel}} = C_1 + C_2 + \cdots$	Capacitors in parallel: add.
$\dfrac{1}{C_{\text{series}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots$	Capacitors in series: reciprocals add.
$E = V/d$	Uniform field between plates. $E$ in V/m (= N/C).