Comprehensive Notes – Capacitors, Current, Resistivity & Ohm’s Law
Capacitors in Series: Core Rules
- When capacitors share a single conducting path (no junctions), they are in series.
- Charge movement chain:
- Charge +q is forced onto the first plate of C_1.
- The same magnitude charge -q is induced on the opposite plate, which in turn becomes +q on the next capacitor, and so on.
- RESULT: All series capacitors carry identical charge q{1}=q{2}=\dots=q{n}=q{eq}.
- Potential (voltage) drops distribute:
- Total battery voltage equals sum of individual drops.
- \Delta V{eq}=\sum{i=1}^{n}\Delta V_i
- Equivalent capacitance for capacitors in series:
\frac{1}{C{eq}}=\sum{i=1}^{n}\frac{1}{Ci}\quad\Rightarrow\quad C{eq}=\Big(\sum \frac{1}{C_i}\Big)^{-1} - Physical analogy: Placing two identical plate capacitors in series is like making one larger capacitor with double the separation d, hence lower C (because C\propto 1/d).
Worked Example – 3 µF, 6 µF, 12 µF, 24 µF in Series (Battery 18 V)
- Identify: single loop → series.
- Compute C{eq}:
\frac{1}{C{eq}}=3^{-1}+6^{-1}+12^{-1}+24^{-1}=0.331\;\mu F^{-1}
C_{eq}=\frac{1}{0.625}=1.6\;\mu F (calculator showed 1/0.625 = 1.6)
- Check: 1.6 µF is smaller than each original value → consistency.
- Charge on the equivalent capacitor:
q=C_{eq}V=(1.6\,\mu F)(18\,\text{V})=28.8\,\mu C
- Series rule → every capacitor gets 28.8\,\mu C.
- Individual voltages using V=q/C:
- V_1=\frac{28.8}{3}=9.6\,\text{V}
- V_2=\frac{28.8}{6}=4.8\,\text{V}
- V_3=\frac{28.8}{12}=2.4\,\text{V}
- V_4=\frac{28.8}{24}=1.2\,\text{V}
- Check sum: 9.6+4.8+2.4+1.2=18\,\text{V} ✓
- Qualitative takeaway: smaller C → larger V in series such that q=CV holds for each.
Mixed Series–Parallel Example (4 µF, 2 µF, 1.5 µF & 12 V Battery)
- Step-by-step reduction stressed (pause and redraw method).
- First identification: 2 µF and 1.5 µF share both nodes → parallel.
- C_{(2\parallel1.5)} = 2+1.5 = 3.5\,\mu F
- New circuit: 4 µF series with 3.5 µF.
- \frac{1}{C{eq}}=4^{-1}+3.5^{-1}\Rightarrow C{eq}=1.867\,\mu F\approx1.9\,\mu F (2 sig-fig)
Charges & Voltages (Back-substitution)
- Equivalent charge:
q=C_{eq}V=(1.867\,\mu F)(12\,\text{V})=22.4\,\mu C - Series part (4 µF vs combined 3.5 µF): share the same charge 22.4 µC.
- V_{4\,\mu F}=\frac{22.4}{4}=5.6\,\text{V}
- Remaining voltage on parallel combo:
V_{3.5\,\mu F}=12-5.6=6.4\,\text{V}
- Parallel rule → both inner capacitors have 6.4 V.
- q_{2\,\mu F}=2\,\mu F\times6.4\,\text{V}=12.8\,\mu C
- q_{1.5\,\mu F}=1.5\,\mu F\times6.4\,\text{V}=9.6\,\mu C
- Check: 12.8+9.6=22.4\,\mu C ✓
- Final answers:
- Charges: 22.4 µC (4 µF), 12.8 µC (2 µF), 9.6 µC (1.5 µF).
- Voltages: 5.6 V, 6.4 V, 6.4 V respectively.
- Pedagogical note: constantly verify with conservation rules (voltages in series sum, charges in parallel sum).
Energy Stored in a Capacitor
- Physical process: moving incremental dq from one plate to the other through potential V(q).
- Work element: dU = V(q)\,dq.
- Integrate (linear V=\frac{q}{C}) from 0→Q:
U = \int_0^{Q} \frac{q}{C}\,dq = \frac{1}{2}\frac{Q^2}{C} - Alternate interchangeable forms via Q=CV:
U=\frac{1}{2}Q V = \frac{1}{2}C V^2 = \frac{1}{2}\frac{Q^2}{C} - Graphical justification: area under a straight-line V(q) plot ⇒ right triangle \frac{1}{2}QV.
- Practical insight: only half the battery’s chemical work becomes stored electrical energy; the other half is dissipated (usually as heat) during charging.
Introduction to Electric Current
- Definition: rate of charge passage through a cross-section.
I=\frac{\Delta Q}{\Delta t} - Unit: ampere (A) = \text{Coulomb}\,/\,\text{second}.
- Quick example: Electric eel moves 2\,\text{mC} in 2\,\text{ms} ⇒ I=1\,\text{A}.
- Empirical indicators of current:
- Wire warms.
- Incandescent bulb glows; brightness ∝ current.
- Compass needle deflects (magnetic field from current).
- Direction convention: Conventional current = flow of positive charge (opposite electron motion).
Conservation of Current (Kirchhoff Junction Rule)
- At any junction: \sum I{in}=\sum I{out}.
- Consequence of charge conservation (no creation/annihilation in wires).
Microscopic Picture: Drift vs Signal
- Electrons undergo frequent collisions with lattice ions; average drift velocity very small \approx10^{-4}\,\text{m/s}.
- Electric field acceleration: a=\frac{eE}{me}, with average time between collisions \tau ⇒ vd=\frac{eE\tau}{2m_e}.
- Electrical signal (field change) propagates near 10^{8}\,\text{m/s} ((\sim c/3)); akin to incompressible water shove.
- Water-hose analogy: pump start moves water everywhere almost simultaneously even though molecules drift slowly.
Resistivity \rho – Material Property
- Defines how strongly a material opposes current independent of geometry.
- Units: \Omega\,m.
- Categories (order-of-magnitude):
- Conductors: \rho\sim10^{-8}\,\Omega m (Cu, Ag).
- Semiconductors: 10^{-5}!\text{–}!10^{-3}\,\Omega m (Si, Ge); \rho drops when heated (negative \alpha).
- Insulators: 10^{8}\,\Omega m and up (glass, rubber).
- Temperature dependence:
\rho(T)=\rho0[1+\alpha (T-T0)]
- \alpha = temperature coefficient (positive for metals, negative for semiconductors).
Resistance R of a Conductor
- Geometry + material:
R=\rho\,\frac{L}{A}
- L = length (m)
- A = cross-section (m^2). For circular wire A=\pi r^2=\pi d^2/4.
- Temperature also: R(T)=R0[1+\alpha (T-T0)].
- Units: ohm (Ω) where 1\,\Omega=1\,\text{V}/\text{A}.
Batteries, EMF & Ohm’s Law Revisited
- Battery maintains nearly constant terminal voltage (EMF \mathcal{E}) until near depletion.
\mathcal{E}=\frac{W_{chem}}{q} - Ohm’s law for ohmic materials:
\Delta V = I R
- Think independent ((R,\Delta V)) vs dependent ((I)).
- Changing parameters:
- Double \Delta V with same R → double I.
- Thicker/shorter wire (lower R) at same \Delta V → higher I.
- Material switch (Cu→Fe) raises \rho hence lowers I.
Ohmic vs Non-Ohmic Devices
- Ohmic: constant R over operating range (standard resistors, many metals at modest temperatures).
- Non-ohmic examples:
- Incandescent bulb (tungsten R rises steeply with temperature → curved V–I graph).
- Batteries (supply fixed \Delta V not fixed R).
- Capacitors/diodes etc. (current depends non-linearly on voltage or time).
Example – 100 W Incandescent Bulb (Tungsten)
- Data: P=100\,\text{W},\;I=0.83\,\text{A},\;V=120\,\text{V}.
- (a) Filament resistance:
R=\frac{V}{I}=\frac{120}{0.83}=1.45\times10^{2}\,\Omega - (b) Length of 35 µm-diameter tungsten wire operating at ~1500 °C.
- Resistivity at 1500 °C: \rho\approx5\times10^{-7}\,\Omega m.
- Area: A=\pi(d/2)^2=\pi (0.035\times10^{-3})^2/4=9.62\times10^{-10}\,m^2.
- L=R\,A/\rho=145\,\Omega\,\cdot9.62\times10^{-10}/(5\times10^{-7})\approx0.278\,m\,(28\,cm\;\approx11\,in).
- Explains tightly coiled long filament inside bulb.
Conceptual & Real-World Connections
- Capacitor energy math links directly to kinetic → potential energy transfer concept; half energy lost observable as heat.
- Temperature coefficients exploited:
- Tungsten’s large \alpha critical for bulbs that self-heat to incandescence.
- Negative \alpha of semiconductors enables thermistors and temperature sensors.
- Magnetic effects of current motivate later study (Ch. 11): compass deflection demonstration.
- Practical advice for circuit problems: redraw after every simplification; annotate Q, \Delta V at each stage; constantly check conservation rules.
Quick Reference: Key Equations
- Series C: \frac{1}{C{eq}}=\sum\frac{1}{Ci}
- Parallel C: C{eq}=\sum Ci
- Capacitor energy: U=\frac{1}{2}CV^2=\frac{1}{2}QV
- Current: I=\dfrac{\Delta Q}{\Delta t}
- Junction rule: \sum I{in}=\sum I{out}
- Resistivity: \rho(T)=\rho_0[1+\alpha\Delta T]
- Resistance: R=\rho\dfrac{L}{A}, R(T)=R_0[1+\alpha\Delta T]
- Ohm’s law: V=IR
- Power (preview): P=IV=I^2R=\dfrac{V^2}{R} (foundation for example, detailed later).
These notes capture all primary derivations, numeric illustrations, conceptual analogies, and practical strategies discussed in the transcript for exam preparation.