Unit 2: Electric Circuits

17–23% of AP Exam · 5 Topics · ~15 Practice Questions

0% Complete

2.1 Current & Resistance

Electric current is the rate of charge flow through a cross section:

$$I = \frac{dQ}{dt} \quad \text{(amperes, A = C/s)}$$

By convention, current direction is the direction positive charge would flow — opposite to electron motion. Current density: \(\vec{J} = \sigma\vec{E}\) where \(\sigma\) is conductivity.

Resistance quantifies how much a material opposes current flow:

$$R = \frac{\rho L}{A} \quad \text{(ohms, Ω)}$$

where \(\rho\) is resistivity (Ω·m), \(L\) is length, \(A\) is cross-sectional area. Resistivity depends on temperature: \(\rho = \rho_0[1 + \alpha(T - T_0)]\). For metals, \(\alpha > 0\) (resistance increases with heat); for semiconductors, \(\alpha < 0\).

Ohm's Law: \(V = IR\) applies to ohmic materials where R is independent of V. Non-ohmic devices (diodes, filaments) have a nonlinear I-V curve.

Electric Power:

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

\(I^2R\) is the Joule heating rate — energy lost as heat in a resistor.

Internal resistance (r) of real batteries: A real battery has internal resistance r in series with its ideal EMF ε. Terminal voltage: \(V_{\text{terminal}} = \varepsilon - Ir\). When current flows, terminal voltage is LESS than the EMF. Open circuit: \(I=0 \Rightarrow V_{\text{terminal}} = \varepsilon\). Short circuit: \(I_{\text{max}} = \varepsilon / r\).

Common Mistake \(P = IV\) is always true for any device. \(P = I^2R\) and \(P = V^2/R\) only apply to resistors (ohmic devices).
Practice

Q2.1: Wire Resistance

A copper wire (\(\rho = 1.7 \times 10^{-8}\ \Omega\cdot\text{m}\)) has length 2.0 m and diameter 1.0 mm. Find its resistance. If 3.0 A flows through it, what is the voltage drop and power dissipated?

Show Solution
\(A = \pi(5\times10^{-4})^2 = 7.85\times10^{-7}\ \text{m}^2\). \(R = \rho L/A = (1.7\times10^{-8})(2)/(7.85\times10^{-7}) = 0.0433\ \Omega\)
\(V = IR = (3)(0.0433) = 0.13\ \text{V}\). \(P = I^2R = (9)(0.0433) = 0.39\ \text{W}\)
Practice

Q2.2: Temperature Dependence

A tungsten filament has \(R = 20\ \Omega\) at 20°C. Its temperature coefficient \(\alpha = 4.5\times10^{-3}/°\text{C}\). Find R at operating temperature 2500°C.

Show Solution
\(R = R_0[1 + \alpha(T - T_0)] = 20[1 + 4.5\times10^{-3}(2480)] = 20[1 + 11.16]\)
\(R = 243\ \Omega\) — about 12× the room-temperature value.

2.2 Series Circuits

In a series circuit, components share one path — current is the same through every element.

Equivalent resistance:

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

Voltage divider: The voltage across each resistor is proportional to its resistance:

$$V_i = \frac{R_i}{R_{\text{eq}}} V_{\text{total}}$$

Key facts:

  • \(I = I_1 = I_2 = \cdots\) (same current everywhere)
  • \(V_{\text{total}} = V_1 + V_2 + \cdots\) (voltages add)
  • Largest resistor gets largest voltage drop (and dissipates most power)

Kirchhoff's Laws

Kirchhoff's Junction Rule (KCL): At any junction, \(\sum I_{\text{in}} = \sum I_{\text{out}}\) — charge is conserved.

Kirchhoff's Loop Rule (KVL): Around any closed loop, \(\sum \Delta V_i = 0\) — energy is conserved. The sum of all voltage rises (EMFs) equals the sum of all voltage drops (IR).

Sign conventions for loop rule: Going across a resistor in the direction of current: −IR. Going across a battery from − to +: +ε. From + to −: −ε.

Application strategy: (1) Label all currents. (2) Write junction equations. (3) Write loop equations for independent loops. (4) Solve the system.

Practice

Q2.3: Voltage Divider

Three resistors (\(R_1=2\ \Omega\), \(R_2=3\ \Omega\), \(R_3=5\ \Omega\)) in series across a 20 V battery. Find the current and the voltage across \(R_2\).

Show Solution
\(R_{\text{eq}} = 2 + 3 + 5 = 10\ \Omega\). \(I = V/R_{\text{eq}} = 20/10 = 2.0\ \text{A}\)
\(V_2 = I R_2 = (2)(3) = 6.0\ \text{V}\). Check: \(V_1 = 4\ \text{V}, V_3 = 10\ \text{V}\), sum = 20 V.

2.3 Parallel Circuits

In a parallel circuit, components share two nodes — voltage is the same across every branch.

Equivalent resistance:

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

For two resistors: \(R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}\)

Current divider: In a two-resistor parallel branch:

$$I_1 = \frac{R_2}{R_1 + R_2} I_{\text{total}}, \quad I_2 = \frac{R_1}{R_1 + R_2} I_{\text{total}}$$

Key facts:

  • \(V = V_1 = V_2 = \cdots\) (same voltage across each branch)
  • \(I_{\text{total}} = I_1 + I_2 + \cdots\) (currents add)
  • \(R_{\text{eq}} < \min(R_1, R_2, \dots)\) — adding a parallel path always decreases R_eq
  • Smallest resistor carries largest current (and dissipates most power)
Critical Series: same CURRENT. Parallel: same VOLTAGE. This is the most fundamental circuit concept — mixing them up ruins every analysis.
Practice

Q2.4: Current Divider

\(R_1 = 4\ \Omega\) in parallel with \(R_2 = 12\ \Omega\), the combination is connected to 24 V. Find all currents and power in each.

Show Solution
\(R_{\text{eq}} = \frac{4\cdot12}{16} = 3\ \Omega\). \(I_{\text{tot}} = 24/3 = 8\ \text{A}\).
\(I_1 = \frac{12}{16}(8) = 6\ \text{A}\), \(I_2 = \frac{4}{16}(8) = 2\ \text{A}\). Check: \(V\) across both = 24 V, so \(I_1 = 24/4 = 6\ \text{A}\), \(I_2 = 24/12 = 2\ \text{A}\). Both methods agree.
\(P_1 = I_1^2 R_1 = 36 \cdot 4 = 144\ \text{W}\); \(P_2 = I_2^2 R_2 = 4 \cdot 12 = 48\ \text{W}\). Total: 192 W = \(I_{\text{tot}}^2 R_{\text{eq}}\).
Practice

Q2.5: Series-Parallel Network

\(R_1 = 6\ \Omega\), \(R_2 = 3\ \Omega\) in parallel with each other, then in series with \(R_3 = 10\ \Omega\). Powered by 18 V. Find \(R_{\text{eq}}\), total current, and \(V_3\).

Show Solution
\(R_{12} = \frac{6\cdot3}{9} = 2\ \Omega\). \(R_{\text{eq}} = 2 + 10 = 12\ \Omega\).
\(I_{\text{tot}} = 18/12 = 1.5\ \text{A}\). \(V_3 = I R_3 = 15\ \text{V}\). \(V_{12} = 3\ \text{V}\).

2.4 RC Circuits — Steady State

After a long time (all transients have decayed), a capacitor reaches steady state:

  • Capacitor acts as an open circuit — no current flows through its branch
  • Voltage across capacitor = voltage of its branch: \(V_C = V_{\text{branch}}\)
  • Charge on capacitor: \(Q = C V_C\)
  • Energy stored: \(U = \frac{1}{2}C V_C^2\)

Analysis strategy:

  1. Replace all capacitors with open circuits
  2. Solve the resulting resistor-network for voltages at capacitor positions
  3. Then \(Q = CV\) and \(U = \frac{1}{2}CV^2\) for each capacitor
Exam Tip "After a long time" = steady state = capacitors are open = no current in capacitor branches. Always redraw the circuit with the open gaps.
Practice

Q2.6: Steady State RC

A 10 V battery, \(R_1 = 2\ \text{k}\Omega\), and \(C = 5\ \mu\text{F}\) in series. A second resistor \(R_2 = 3\ \text{k}\Omega\) is in parallel with C. After a long time, find \(V_C\), \(Q\), U, and the current through each resistor.

Show Solution
At steady state, C is open. Current only in \(R_1 \to R_2\) series path: \(I = 10/(2000+3000) = 2.0\ \text{mA}\).
\(V_C = V_{R_2} = IR_2 = (0.002)(3000) = 6\ \text{V}\). \(Q = CV_C = (5\times10^{-6})(6) = 30\ \mu\text{C}\).
\(U = \frac{1}{2}CV^2 = 9\times10^{-5}\ \text{J}\). \(I_{R_1} = I_{R_2} = 2\ \text{mA}\).

2.5 RC Circuits — Transient Analysis

During charging/discharging, the capacitor voltage and current are exponential functions of time.

Time constant:

$$\tau = RC$$

After \(1\tau\): ~63% of change. After \(5\tau\): effectively at steady state (>99%).

Charging (RC series, battery ε):

$$q(t) = Q_f\left(1 - e^{-t/RC}\right), \quad Q_f = C\varepsilon$$ $$V_C(t) = \varepsilon\left(1 - e^{-t/RC}\right)$$ $$I(t) = \frac{\varepsilon}{R}e^{-t/RC} = I_0 e^{-t/RC}$$

Discharging (no battery):

$$q(t) = Q_0 e^{-t/RC}$$ $$V_C(t) = V_0 e^{-t/RC}$$ $$I(t) = \frac{V_0}{R}e^{-t/RC} = I_0 e^{-t/RC}$$

Standard derivative forms (memorize):

  • Charging: \(\frac{dq}{dt} = \frac{\varepsilon}{R}e^{-t/RC}\)
  • At \(t=0\): uncharged capacitor acts like a short (zero voltage)
  • At \(t \to \infty\): capacitor acts like an open (no current)
Common Mistake Writing \(V_C(t) = \varepsilon e^{-t/RC}\) for charging. That's the form for discharging. Charging is: \(V_C = \varepsilon(1 - e^{-t/RC})\). Always check: at t=0, your formula should give \(V_C = 0\) for charging, \(V_C = V_0\) for discharging.
Practice

Q2.7: Charging — Time to 90%

An uncharged 2.0 μF capacitor and 50 kΩ resistor in series with a 12 V battery. Find: (a) time constant, (b) time for V_C to reach 10.8 V (90% of final), (c) current at that time.

Show Solution
(a) \(\tau = RC = (5\times10^4)(2\times10^{-6}) = 0.10\ \text{s}\)
(b) \(V_C = 12(1 - e^{-t/0.1}) = 10.8 \implies 1 - e^{-t/0.1} = 0.9 \implies e^{-t/0.1} = 0.1 \implies t = -0.1\ln(0.1) = 0.230\ \text{s}\)
(c) At that time, \(I = (\varepsilon/R)e^{-t/RC} = (12/50000)e^{-2.303} = (2.4\times10^{-4})(0.1) = 24\ \mu\text{A}\)

2.6 Meters & Measurement

Ammeters measure current. They must be connected in series and have very low internal resistance (ideally zero) so they don't affect the circuit.

Voltmeters measure potential difference. They must be connected in parallel across the component and have very high internal resistance (ideally infinite) so they draw negligible current.

Practice

Q2.8: Battery with Internal Resistance

A battery with ε = 9.0 V and internal resistance r = 0.50 Ω is connected to a 4.0 Ω resistor. Find: (a) circuit current, (b) terminal voltage of the battery, (c) power delivered to the load, (d) power dissipated inside the battery.

Show Solution
\(I = \varepsilon/(R+r) = 9/(4+0.5) = 2.0\ \text{A}\).
\(V_{\text{terminal}} = \varepsilon - Ir = 9 - (2)(0.5) = 8.0\ \text{V}\) (or \(V_{\text{terminal}} = IR = 2\times4 = 8\ \text{V}\)).
\(P_{\text{load}} = I^2R = 16\ \text{W}\). \(P_{\text{internal}} = I^2r = 2.0\ \text{W}\). Total: 18 W = εI.
Practice

Q2.9: Two-Loop Circuit (Kirchhoff)

Two batteries: ε₁ = 12 V, ε₂ = 6 V. Resistors: R₁ = 4 Ω connected to ε₁, R₂ = 3 Ω connected to ε₂, R₃ = 2 Ω connects the midpoints. Find all three branch currents.

Show Solution
Junction (top): \(I_1 = I_2 + I_3\).
Left loop: \(12 - 4I_1 - 2I_3 = 0 \Rightarrow 4I_1 + 2I_3 = 12\).
Right loop: \(-6 + 3I_2 - 2I_3 = 0 \Rightarrow 3I_2 - 2I_3 = 6\).
Substitute \(I_2 = I_1 - I_3\) into right loop: \(3(I_1 - I_3) - 2I_3 = 6 \Rightarrow 3I_1 - 5I_3 = 6\).
Solve system: \(I_1 = 2.0\ \text{A}\), \(I_3 = 0\ \text{A}\), \(I_2 = 2.0\ \text{A}\). The midpoint resistor carries no current.