Current Electricity — Notes

Current $I = dq/dt$ (ampere, A = C s$^{-1}$); conventional direction = direction of positive-charge flow. In metals, electrons move opposite to it.

Drift velocity. Free electrons move randomly (no net velocity without a field). In a field each gains $-e\vec E/m$ between collisions; averaged over the relaxation time $\tau$: $\vec v_d = -\dfrac{e\vec E}{m}\tau$

Current–drift relation. In time $\Delta t$ the electrons in a length $v_d\Delta t$ cross area $A$: $\Delta q = n e A v_d\Delta t$, so $I = n e A v_d,\qquad j = \dfrac{I}{A} = n e v_d$

Mobility $\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m}$ (m² V$^{-1}$ s$^{-1}$).

Typical drift speed: about $10^{-4}$ m s$^{-1}$; the field (and so the signal) travels at nearly the speed of light, which is why a lamp lights at once.

Substituting $v_d = eE\tau/m$ and $E = V/l$ into $I = n e A v_d$: $V = IR,\qquad R = \rho\dfrac{l}{A},\qquad \rho = \dfrac{m}{n e^2\tau},\qquad \sigma = \dfrac{1}{\rho},\qquad \vec j = \sigma\vec E$

• $R$ (ohm) depends on the sample; $\rho$ (Ω m) only on the material and temperature.
• Stretching a wire $k$ times (constant volume): $R \to k^2R$.

V–I characteristics. Linear (ohmic): straight line through the origin, same for both directions. Non-linear: filament lamp (R rises as it heats), diode (depends on the sign of V), GaAs (more than one V for some I).

Temperature dependence.

• Metals: $\rho_T = \rho_0[1 + \alpha(T - T_0)]$, $\alpha > 0$ — hotter ions, more collisions, smaller $\tau$.
• Alloys (nichrome, manganin, constantan): very small $\alpha$ — standard resistors and heater elements.
• Semiconductors: $\rho$ falls with temperature ($\alpha < 0$) because $n$ rises steeply.

Charge $q$ moving through potential difference $V$ transfers energy $qV$. Power delivered to a resistor: $P = VI = I^2R = \dfrac{V^2}{R}$

• Series (same $I$): the larger $R$ dissipates more ($I^2R$).
• Parallel (same $V$): the smaller $R$ dissipates more ($V^2/R$).
• Rated values give $R = V_\text{rated}^2/P_\text{rated}$; then use the actual voltage.
• Energy $W = Pt$; 1 kWh $= 3.6 \times 10^{6}$ J.
• Power is sent over long distances at high voltage so that the current, and the $I^2R$ loss in the lines, is small.

EMF $E$: terminal p.d. of a cell when it supplies no current. Internal resistance $r$: resistance of the electrolyte and electrodes.

• Discharging: $V = E - Ir$, $I = \dfrac{E}{R + r}$. Charging: $V = E + Ir$.
• Short circuit: $I = E/r$. Open circuit: $V = E$.
• $V$–$I$ graph: straight line, intercept $E$, slope $-r$.

Combinations

• Series: $E_\text{eq} = \sum E_i$ (a reversed cell subtracts), $r_\text{eq} = \sum r_i$.
• Parallel: $\dfrac{E_\text{eq}}{r_\text{eq}} = \sum \dfrac{E_i}{r_i}$, $\dfrac{1}{r_\text{eq}} = \sum \dfrac{1}{r_i}$. For $n$ identical cells: $E$, $r/n$.

Junction rule (charge conservation): $\sum I_\text{in} = \sum I_\text{out}$ at every junction.

Loop rule (energy conservation): $\sum \Delta V = 0$ round any closed loop.

Method: (1) mark a current in every branch with an assumed direction; (2) apply the junction rule to reduce unknowns; (3) choose independent loops and add potential changes with the sign rules (resistor along I: $-IR$; cell − to +: $+E$); (4) solve; a negative current flows opposite to the arrow.

Worked result (deck example): 12 V/2 Ω and 6 V/2 Ω branches with 4 Ω across them give 2.4 A, −0.6 A (the 6 V cell is charged) and 1.8 A.

Arms P (A–B), Q (B–C), R (A–D), S (D–C); cell across A–C, galvanometer across B–D.

Balance ($I_g = 0$): B and D at the same potential. Loop ABDA: $I_1P = I_2R$; loop BCDB: $I_1Q = I_2S$. Dividing: $\dfrac{P}{Q} = \dfrac{R}{S}$ Balance is independent of the emf and of the galvanometer’s resistance; the cell and galvanometer may be interchanged.

Metre bridge: two arms are lengths $l$ and $100 - l$ (cm) of a uniform wire: $\dfrac{X}{R} = \dfrac{l}{100 - l}$ (X in the left gap). Best accuracy with balance near the middle; end errors are reduced by interchanging X and R and averaging.