Current Electricity

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Current Electricity — Key Facts

Electric Current: Current $I = Q/t$ = rate of flow of charge. For a uniform conductor: $I = nAev_d$ where $n$ = charge carrier density (per m³), $A$ = cross-section, $e = 1.6 × 10^{-19}$ C, $v_d$ = drift velocity.

Current density $\vec{J} = n e \vec{v_d} = \sigma \vec{E}$ (Ohm’s law in vector form), where $\sigma = 1/\rho$ is conductivity.

Resistance: $R = \rho L/A$. Temperature dependence: $\rho_T = \rho_0[1 + \alpha(T - T_0)]$. For conductors: $\alpha > 0$ (resistance increases with $T$). For semiconductors: $\alpha < 0$ (resistance decreases with $T$ — carrier concentration increases faster than scattering).

Series and Parallel: Series: $R_{eq} = R_1 + R_2 + …$ (same current, voltages add). Parallel: $1/R_{eq} = 1/R_1 + 1/R_2 + …$ (same voltage, currents add). Two resistors $R_1, R_2$ in parallel: $R_{eq} = R_1 R_2/(R_1 + R_2)$.

⚡ Exam tip: In series, the largest resistor drops the most voltage ($V = IR$). In parallel, the smallest resistor carries the most current.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Current Electricity — CUET Study Guide

EMF and Internal Resistance: For a real source with emf $\varepsilon$ and internal resistance $r$, terminal voltage $V = \varepsilon - Ir$. When delivering maximum power to an external load $R$: $P = I^2R = (\varepsilon/(R+r))^2 × R$. $dP/dR = 0$ when $R = r$. Maximum power $P_{max} = \varepsilon^2/(4r)$.

Kirchhoff’s Laws:

• Junction law: $\sum I_{in} = \sum I_{out}$ (charge conservation)
• Loop law: $\sum \varepsilon = \sum IR$ around any closed loop (energy conservation)

Wheatstone Bridge: Balanced when $R_1/R_2 = R_3/R_4$. The bridge resistor (connecting the two potential-divider midpoints) carries zero current and can be ignored for equivalent resistance calculation.

For unbalanced bridge: the equivalent resistance requires Delta-Star transformation or solving using Kirchhoff’s laws.

Metre Bridge (Slide Wire Bridge): A uniform wire of length 100 cm acts as one resistance arm. Balance condition: $R_1/R_2 = l_1/l_2$. Unknown resistance $X = R \times l_2/l_1$ where $R$ is the known resistance and $l_1, l_2$ are the lengths of wire on either side of the balance point.

Potentiometer: Measures emf without drawing current. Working: a steady current $I$ flows through the potentiometer wire. The potential drop per unit length $k = IR/L$ is constant. Unknown emf $\varepsilon_x$ balances at length $l_x$: $\varepsilon_x = kl_x$.

Comparison of emfs: $\varepsilon_1/\varepsilon_2 = l_1/l_2$. Internal resistance: With switch open, balance at $l_1$: $\varepsilon = kl_1$. With switch closed (load $R$), balance at $l_2$: $V = kl_2$. Internal resistance $r = (\varepsilon/V - 1)R = (l_1/l_2 - 1)R$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Current Electricity — Comprehensive Physics Notes

Drift Velocity — Detailed: Electrons in a metal move randomly at thermal velocity $v_{th} ≈ 10^6$ m/s at room temperature. When an electric field $E$ is applied, a small net drift is superimposed. $v_d = \mu E$ where $\mu$ = electron mobility (m²/V·s). For copper at 300 K: electron density $n = 8.5 × 10^{28}$ m⁻³, $\mu = 4.5 × 10^{-3}$ m²/V·s. For $E = 0.1$ V/m (in household wiring): $v_d ≈ 4.5 × 10^{-4}$ m/s = 0.45 mm/s. Current appears instantaneous because the electric field propagates at near $c$.

Derivation of $I = nAev_d$: Consider a conductor of cross-section $A$. In time $\Delta t$, electrons travel $v_d \Delta t$ in the field direction. All electrons within volume $A \cdot v_d \Delta t$ will pass through the cross-section. Number of electrons = $n A v_d \Delta t$. Charge flowing = $n A v_d \Delta t × e$. $I = Q/\Delta t = nAev_d$.

Kirchhoff’s Laws — Matrix Method: For a circuit with $n$ independent loops:

1. Assign mesh currents $I_1, I_2, …, I_n$ (clockwise for each mesh)
2. Write $\sum V = 0$ around each mesh: sum of emfs = sum of $RI$ drops
3. Solve the system of equations

Example: Two meshes with shared resistor $R_3$. Mesh 1: $\varepsilon_1 = I_1 R_1 + (I_1 - I_2)R_3$. Mesh 2: $\varepsilon_2 = I_2 R_2 + (I_2 - I_1)R_3$. Solve simultaneously for $I_1$ and $I_2$.

Delta-Star Transformation: Delta ($R_{AB}, R_{BC}, R_{CA}$) ↔ Star ($R_A, R_B, R_C$): $R_A = \frac{R_{AB} R_{CA}}{R_{AB} + R_{BC} + R_{CA}}$, cyclic permutations.

For equal resistors $R$ in delta: each star arm = $R/3$. For equal resistors $R$ in star: delta resistor = $3R$.

Wheatstone Bridge — Sensitivity: A Wheatstone bridge is most sensitive when all four arms have resistances of the same order. The galvanometer detects the smallest unbalance. For maximum sensitivity, the bridge should be in its balanced condition initially (or nearly so).

Potentiometer — Applications:

1. Measuring unknown emf (compare with standard cell)
2. Measuring internal resistance of a cell
3. Comparing resistances
4. Calibrating voltmeters and ammeters

Temperature Coefficient of Resistance: For metals: $R_T = R_0[1 + \alpha(T - T_0)]$. For semiconductors: $R_T = R_0 e^{E_g/(2kT)}$ where $E_g$ = band gap, $k$ = Boltzmann constant.

The temperature coefficient $\alpha$ for copper = $3.93 × 10^{-3}$ °C⁻¹. For nichrome = $0.0004$ °C⁻¹ (nearly constant — used for standard resistors).

Thermoelectric Effects:

• Seebeck effect: Two dissimilar metals joined at two junctions — a temperature difference produces an emf. Used in thermocouples. Seebeck coefficient $S = d\varepsilon/dT$.
• Peltier effect: Passing current through a junction of two metals produces heating or cooling. Reversible.
• Thomson effect: Temperature gradient along a single conductor produces an emf.

Thermocouple calibration: Type K (chromel-alumel): ~40 μV/°C, range -200°C to +1200°C. Type T (copper-constantan): ~40 μV/°C, range -250°C to +350°C.

JEE Pattern Analysis: CUET/JEE questions frequently test: (1) Meter bridge for unknown resistance, (2) Potentiometer for comparing emfs and measuring internal resistance, (3) Kirchhoff’s laws for complex networks, (4) Drift velocity and current density relationship, (5) Maximum power transfer theorem. JEE 2023: “In a potentiometer, the balance point is found at 60 cm when a cell of emf 1.5 V is used. With another cell of unknown emf, the balance point is at 80 cm. Find the unknown emf.” Answer: $\varepsilon_2/\varepsilon_1 = 80/60$. $\varepsilon_2 = 1.5 × 80/60 = 2.0$ V.

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