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 $q<em>{1}=q</em>{2}=\dots=q<em>{n}=q</em>{eq}$.
• Potential (voltage) drops distribute:
  • Total battery voltage equals sum of individual drops.
  • $\Delta V<em>{eq}=\sum</em>{i=1}^{n}\Delta V_i$
• Equivalent capacitance for capacitors in series:
  $\frac{1}{C<em>{eq}}=\sum</em>{i=1}^{n}\frac{1}{C<em>i}\quad\Rightarrow\quad C</em>{eq}=\Big(\sum \frac{1}{C_i}\Big)^{-1}$
• 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 $C<em>{eq}$: $\frac{1}{C</em>{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).

1. First identification: 2 µF and 1.5 µF share both nodes → parallel.
   • $C_{(2\parallel1.5)} = 2+1.5 = 3.5\,\mu F$
2. New circuit: 4 µF series with 3.5 µF.
   • $\frac{1}{C<em>{eq}}=4^{-1}+3.5^{-1}\Rightarrow C</em>{eq}=1.867\,\mu F\approx1.9\,\mu F$ (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: $\sum I<em>{in}=\sum I</em>{out}$.
• 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: $a=\frac{eE}{m<em>e}$, with average time between collisions $\tau$ ⇒ $v</em>d=\frac{eE\tau}{2m_e}$.
• 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: $\rho(T)=\rho<em>0[1+\alpha (T-T</em>0)]$
  • $\alpha$ = 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: $R(T)=R<em>0[1+\alpha (T-T</em>0)]$.
• 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 $C$: $\frac{1}{C<em>{eq}}=\sum\frac{1}{C</em>i}$
• Parallel $C$: $C<em>{eq}=\sum C</em>i$
• Capacitor energy: $U=\frac{1}{2}CV^2=\frac{1}{2}QV$
• Current: $I=\dfrac{\Delta Q}{\Delta t}$
• Junction rule: $\sum I<em>{in}=\sum I</em>{out}$
• 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.