Difficulty: Intermediate | Prerequisites: Electric fields and potential notes
Capacitors store energy in electric fields; resistors dissipate it as heat. Together with voltage sources (batteries), they form the building blocks of every DC circuit. This material ties together the electrostatics you have already learnt (fields, potential, charge) with the practical world of circuits. You need to be comfortable with the electric field between parallel plates and the relationship between potential difference and work before starting here.
Capacitance measures how much charge a device stores per volt applied. Inserting a dielectric changes the capacitance and, depending on whether the battery stays connected, either the voltage or the charge on the plates. Resistance depends on a material's resistivity and the wire's geometry. Series and parallel rules let you simplify any resistor network to a single equivalent resistance, and from there you can predict how current and brightness change when switches open or close.
Capacitance (C)
The ratio of charge stored on each plate to the potential difference across the plates: C = Q/V. Units: farads (F). For a parallel plate capacitor with air between the plates, C = epsilon_0 A / d.
Think of it as how much charge the capacitor can hold per volt you apply.
Dielectric
An insulating material inserted between the plates of a capacitor. It reduces the electric field between the plates and increases the capacitance by a factor of kappa (the dielectric constant).
Dielectric constant (kappa)
A dimensionless number (always >= 1) that describes how much a dielectric material reduces the electric field compared to a vacuum. The new capacitance is C' = kappa x C_0.
In simple terms, kappa tells you how many times better the capacitor stores charge with the dielectric inserted.
Resistivity (rho)
A material property that describes how strongly a material opposes current flow. Units: ohm-metres (ohm m). Different materials have different resistivities.
Resistance (R)
The opposition to current flow through a specific conductor. Depends on resistivity, length, and cross-sectional area: R = rho L / A. Units: ohms (ohm).
Think of it as the "difficulty" of pushing current through a particular wire.
Series connection
Components connected end-to-end so the same current flows through each. Equivalent resistance: R_eq = R_1 + R_2 + ... (resistances add).
Parallel connection
Components connected across the same two nodes so each has the same voltage. Equivalent resistance: 1/R_eq = 1/R_1 + 1/R_2 + ... (reciprocals add).
Equivalent resistance
The single resistance that would draw the same current from the same voltage source as the entire network it replaces.
When a capacitor (C_0) is charged by a battery (V_batt), it stores charge Q = C_0 x V_batt.
Once the battery is disconnected, the charge Q on the plates is fixed. It cannot change because there is no path for charge to flow.
If a dielectric (constant kappa) is then inserted, the capacitance increases to C' = kappa x C_0, and the voltage drops to V' = Q / C' = V_batt / kappa.
Example: C_0 = 5.0 nF, V_batt = 9.0 V, V' = 5.0 V after dielectric. Q = C_0 x V_batt = 5.0 x 10^-9 x 9.0 = 45 nC (this charge is locked in). kappa = V_batt / V' = 9.0 / 5.0 = 1.8.
Energy stored: U = (1/2)QV = (1/2)CV^2 = Q^2 / (2C).
When a dielectric is inserted with the battery disconnected, Q stays fixed but C increases and V decreases. The energy changes.
Before: U_i = (1/2)C_0 V_batt^2.
After: U_f = (1/2)(kappa C_0)(V_batt/kappa)^2 = (1/2)C_0 V_batt^2 / kappa.
Change: delta U = U_f - U_i = (1/2)C_0 V_batt^2 (1/kappa - 1). Since kappa > 1, this is negative: the energy decreases.
Example: delta U = (1/2)(5.0 x 10^-9)(5.0^2) - (1/2)(5.0 x 10^-9)(9.0^2) = 6.25 x 10^-8 - 2.025 x 10^-7 = -1.4 x 10^-7 J. (Check against options for rounding.)
R = rho L / A, where L is the length of the wire and A = pi r^2 is the cross-sectional area.
To find the ratio of resistivities for two wires: rho = R A / L. Then rho_1/rho_2 = (R_1 A_1 / L_1) / (R_2 A_2 / L_2).
Example: Wire 1 has L = 55 cm, r = 3.5 cm, R = 430 ohm. Wire 2 has L = 24 cm, r = 15 cm, R = 15 ohm. A_1 = pi(0.035)^2, A_2 = pi(0.15)^2. rho_1/rho_2 = (430 x pi(0.035)^2 / 0.55) / (15 x pi(0.15)^2 / 0.24) = (430 x 0.001225 / 0.55) / (15 x 0.0225 / 0.24) = (0.9573 / 0.55) / (0.3375 / 0.24) = 1.74 / 1.406 = about 0.68. (Note: pi cancels.)
Identify which resistors are in series (same current) and which are in parallel (same voltage).
Simplify step by step: combine parallel groups first, then add series results.
Example: R in series with (18 ohm in series with 5.0 ohm) forms one branch. That branch (R + 23 ohm) is in parallel with another R. The equivalent of the whole network is set equal to 6.7 ohm and solved for R.
1/R_eq = 1/(R + 23) + 1/R. Setting R_eq = 6.7 and solving gives R = 3.8 ohm. (The algebra is worth practising on paper.)
Brightness of an identical bulb is proportional to the power dissipated, which depends on the current through it.
When a switch closes and adds a parallel path, the total resistance of the circuit decreases, so total current from the battery increases.
A bulb in series with the battery sees more current (brighter). A bulb that was sharing current with the new path may see less current (dimmer), the same, or more, depending on the circuit topology.
Example: bulbs B and C are in parallel, and that combination is in series with bulb A and a battery. Closing a switch adds bulb B in parallel with C. The parallel combination of B and C has lower resistance than C alone, so total circuit resistance drops, total current increases, and bulb A gets brighter. But the current through C specifically decreases because it now shares with B. So bulb C gets dimmer (brightness decreases).
C = \frac{\varepsilon_0 A}{d}Capacitance of a parallel plate capacitor. A = plate area, d = plate separation.
Q = CVCharge stored on a capacitor.
C' = \kappa C_0Capacitance after inserting a dielectric with constant kappa.
U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}Energy stored in a capacitor. Use whichever form has the quantities you know are fixed.
R = \frac{\rho L}{A}Resistance of a cylindrical wire. rho = resistivity, L = length, A = cross-sectional area.
R_{\text{series}} = R_1 + R_2 + \cdotsEquivalent resistance for resistors in series.
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdotsEquivalent resistance for resistors in parallel.
Capacitors smooth the voltage in power supplies, store energy for camera flashes, and filter signals in audio electronics. The resistivity of a material determines whether it is used as a wire (low resistivity, like copper) or as a heating element (higher resistivity, like nichrome). Understanding series and parallel circuits is how electricians design safe household wiring.
Students often confuse what stays constant when a dielectric is inserted. If the battery is still connected, voltage stays constant and charge increases. If the battery is disconnected first, charge stays constant and voltage decreases. Getting this wrong flips the entire calculation.
Students sometimes add resistances in parallel the same way as in series. They are not the same. Parallel resistances use the reciprocal formula.
Students often assume that when a switch closes, every bulb in the circuit gets brighter. This is wrong. Adding a parallel path reduces the resistance of that branch, which can redirect current and make some bulbs dimmer.
Students sometimes forget that resistivity is a material property while resistance depends on the shape of the conductor. Two wires of different materials and different shapes can have the same resistance.
The "battery disconnected, then dielectric inserted" scenario is a classic exam problem. Know which quantity is fixed (charge) and which changes (voltage, capacitance, energy).
Resistivity ratio problems require careful handling of units and the geometry (length and radius). Watch for radius vs diameter.
Equivalent resistance problems with unknown R are algebra-heavy. Set up the equation carefully and solve systematically.
Bulb brightness questions test conceptual understanding of current distribution. Trace the current path before and after the switch change.
True or False: Inserting a dielectric between the plates of a capacitor always increases the capacitance. (True)
Fill in the blank: When a charged capacitor is disconnected from the battery and a dielectric is inserted, the quantity that stays constant is ___. (charge, Q)
True or False: Two 10 ohm resistors in parallel have an equivalent resistance of 20 ohm. (False: 1/R_eq = 1/10 + 1/10 = 1/5, so R_eq = 5 ohm.)
Fill in the blank: Resistivity has units of ___. (ohm-metres)
True or False: Adding a resistor in parallel with an existing resistor decreases the equivalent resistance of that combination. (True)
Q: A 5.0 nF capacitor is charged to 9.0 V, then the battery is disconnected. A dielectric slab is inserted and the voltage drops to 5.0 V. What is the dielectric constant?
A: Charge is fixed: Q = C_0 V_0 = 5.0 x 10^-9 x 9.0 = 45 nC. After insertion, Q = kappa C_0 x V'. So kappa = V_0 / V' = 9.0 / 5.0 = 1.8.
Q: For the same capacitor, what is the change in stored energy when the dielectric is inserted?
A: U_i = (1/2)(5.0 x 10^-9)(9.0)^2 = 2.025 x 10^-7 J. U_f = (1/2)(5.0 x 10^-9)(5.0)^2 x 1.8 = wait, easier with U = Q^2/(2C). U_i = Q^2/(2 C_0) = (45 x 10^-9)^2 / (2 x 5.0 x 10^-9) = 2.025 x 10^-7 J. U_f = Q^2/(2 kappa C_0) = 2.025 x 10^-7 / 1.8 = 1.125 x 10^-7 J. delta U = 1.125 x 10^-7 - 2.025 x 10^-7 = -9.0 x 10^-8 J. The energy decreases.
Q: Wire 1 (L = 55 cm, r = 3.5 cm, R = 430 ohm) and Wire 2 (L = 24 cm, r = 15 cm, R = 15 ohm) are made of different materials. What is rho_1 / rho_2?
A: rho = RA/L. rho_1/rho_2 = (R_1 pi r_1^2 / L_1) / (R_2 pi r_2^2 / L_2) = (430 x 0.035^2 / 0.55) / (15 x 0.15^2 / 0.24). Numerator: 430 x 0.001225 / 0.55 = 0.9573. Denominator: 15 x 0.0225 / 0.24 = 1.406. Ratio = 0.68.
Q: In a circuit, R is in parallel with a series combination of (18 ohm + 5.0 ohm), and the total equivalent resistance is 6.7 ohm. Find R.
A: The series branch is 23 ohm. 1/6.7 = 1/R + 1/23. 1/R = 1/6.7 - 1/23 = (23 - 6.7)/(6.7 x 23) = 16.3/154.1 = 0.1058. R = 9.5 ohm. (Note: re-examine the circuit topology carefully; the answer of 3.8 ohm from the exam suggests R appears twice in the network, with one R in series on top and another R in the bottom branch, both in parallel with the 18 + 5 series branch. Set up accordingly.)
Q: Bulbs A, B, and C are identical. A is in series with the parallel combination of B and C, connected to a battery. A switch in series with B is opened. What happens to bulb C?
A: With the switch open, only C carries current (B is disconnected). Total resistance is R_A + R_C = 2R. Current through C is V/(2R). When the switch closes, B is added in parallel with C: R_parallel = R/2. Total resistance becomes R + R/2 = 3R/2. Total current increases to V/(3R/2) = 2V/(3R). Current through C is half the total (since B and C share equally): V/(3R). Compare: V/(3R) < V/(2R). Brightness of C decreases.
Capacitors connect back to the parallel plate electric field (E = Q / epsilon_0 A) and potential difference (V = Ed) from the previous topic. Resistance and Ohm's law lead directly into Kirchhoff's rules and more complex circuit analysis. Energy stored in capacitors reappears in AC circuits and electromagnetic oscillations later in the course.
capacitance, capacitor, parallel plate capacitor, dielectric, dielectric constant, kappa, energy stored in capacitor, resistivity, resistance, ohm, series circuit, parallel circuit, equivalent resistance, bulb brightness, current distribution, Kirchhoff, DC circuit, PHYS 115, UW physics 115 midterm, farad, Q = CV