Current Electricity – JEE Main Physics Formula Sheet & Class 12 Notes | Ohm's Law, Kirchhoff's Laws, Wheatstone Bridge, Potentiometer, RC Circuit & All Formulas

In Current Electricity, drift velocity v_d from Aakash PDF: free electrons in a metal have random thermal motion (very fast, ~10⁵ m/s) but no net displacement. When electric field E is applied, electrons experience force F=eE and undergo net drift opposite to E. Between collisions (average time τ), electron accelerates: v_d=eE/mτ. Current relation: consider cross-section area A, n=electron density. In time dt, electrons within distance v_d dt reach the cross-section → charge dQ=n×(A×v_d×dt)×e. I=dQ/dt=nAev_d → v_d=I/nAe. Typical values: for copper (n=8.5×10²⁸/m³), I=1A, A=1mm²=10⁻⁶m²: v_d=1/(8.5×10²⁸×10⁻⁶×1.6×10⁻¹⁹)≈7.4×10⁻⁵ m/s (very slow!). Yet signal travels at ~c (speed of light) because E field propagates at c. JEE Main Current Electricity: "if cross-section doubles, I same, find v_d ratio" → v_d=I/nAe → v_d∝1/A → v_d halves. "If temperature doubles (n≈const, τ decreases), find ρ ratio" → ρ=m/ne²τ → ρ∝1/τ → ρ increases.

In Current Electricity, EMF ε is the total energy per unit charge provided by the cell (voltage when no current flows). Internal resistance r causes a voltage drop Ir when current I flows. Terminal voltage V=voltage measured at cell terminals: V=ε–Ir (cell discharging, current out of positive terminal). V<ε when current is drawn. V=ε when I=0 (open circuit) or r=0 (ideal cell). V=ε+Ir (cell charging, current forced into positive terminal — like during recharging a battery). V>ε during charging. Derivation: for discharge, KVL around loop: ε=Ir+IR → I=ε/(R+r). Terminal V=IR=ε–Ir. As R→∞ (open circuit): I→0, V→ε. As R→0 (short circuit): I→ε/r (max), V→0. For given ε and r, external power P_ext=I²R=ε²R/(R+r)². Maximum when dP/dR=0 → R=r (load resistance = internal resistance) → P_max=ε²/4r. Efficiency at max power = P_ext/(εI)=½×ε²/(R+r)/ε×(R/(R+r))... = R/(R+r)=r/(r+r)=50%. JEE Main: "find R for max power from cell with ε=6V, r=1Ω" → R=r=1Ω; P_max=36/4=9W.

In Current Electricity, Kirchhoff's laws from Aakash PDF: KCL (junction law): at any junction, Σ(currents in)=Σ(currents out). Based on charge conservation. Sign convention: currents entering junction are +; leaving are –. KVL (loop law): around any closed loop, sum of all potential drops=0, OR sum of EMFs = sum of IR drops. Based on energy conservation. Sign convention: (a) Resistor: traverse in direction of assumed current → –IR (voltage drop); traverse opposite → +IR. (b) EMF source: traverse from – to + terminal → +ε (gain); traverse from + to – → –ε (loss). Application method: (1) Assign current directions to each branch (assumed; if result negative, actual direction is opposite). (2) Apply KCL at all junctions (one will be redundant). (3) Choose independent loops and apply KVL to each. (4) Solve simultaneous equations. Number of independent equations: KCL gives (j–1) equations (j junctions); KVL gives (b–j+1) equations (b branches, j junctions). JEE Main Current Electricity: two-loop circuit with two batteries → 3 unknown currents → 2 KCL equations (one junction) + 2 KVL equations (two loops) → solve.

In Current Electricity, Wheatstone bridge from Aakash PDF: four resistors P (top-left), Q (bottom-left), R (top-right), S (bottom-right) in a diamond. Battery between top and bottom nodes (A and C). Galvanometer between left-right midpoints (B and D). At balance: no current in galvanometer (I_g=0). This means V_B=V_D. From the top paths: V_A–V_B=IP (voltage across P) and V_B–V_C=IQ (same current I since no galvanometer current). Similarly V_A–V_D=IR' and V_D–V_C=IS'. At balance I_P=I_Q (same current in left arm) and I_R=I_S (same current in right arm). V_B=V_A–IP=V_A–I_left×P. V_D=V_A–I_right×R. For V_B=V_D: I_left×P=I_right×R. Also V_C=V_B–I_left×Q=V_D–I_right×S. So I_left×Q=I_right×S. Dividing: P/Q=R/S. Equivalently P×S=Q×R. JEE Main Current Electricity: "P=2Ω, Q=3Ω, R=4Ω. Find S for balance" → S=QR/P=3×4/2=6Ω.

In Current Electricity, potentiometer from Aakash PDF: a long uniform wire (length L, total resistance R_wire) connected to a driving cell. A steady current I₀ flows, creating a uniform potential drop along the wire. Potential gradient: φ=I₀×R_wire/L=V_AB/L (V/m or V/cm). Potential drop across length l: V=φl ∝ l. Principle: compare unknown EMF with potential drop across a length l₀ where balance (no current) is achieved. Application 1 — EMF comparison: two cells ε₁ and ε₂ balance at lengths l₁ and l₂ respectively. φl₁=ε₁ and φl₂=ε₂ → ε₁/ε₂=l₁/l₂. Application 2 — Internal resistance: cell EMF ε, internal resistance r, balanced at l₁ (open circuit, I=0 from cell). Now connect R externally; terminal voltage V=ε–Ir balanced at l₂ (l₂

In Current Electricity, RC circuits from Aakash PDF: Charging (capacitor initially uncharged, battery ε connected at t=0): KVL: ε=Q/C+IR → ε=Q/C+(dQ/dt)R. Solution: Q(t)=Cε(1–e^(–t/RC))=Q₀(1–e^(–t/RC)) where Q₀=Cε (final/steady-state charge). Voltage: V_C(t)=ε(1–e^(–t/RC)). Current: I(t)=dQ/dt=(ε/R)e^(–t/RC)=I₀e^(–t/RC) (starts at I₀=ε/R, exponentially decays). Time constant τ=RC (unit: Ω×F=s). At t=τ: Q=Q₀(1–1/e)≈0.632Q₀; V_C≈0.632ε; I≈0.368I₀. At t=5τ: Q≈0.993Q₀ (considered fully charged). Discharging (fully charged Q₀, battery removed, discharges through R at t=0): Q(t)=Q₀e^(–t/RC). V_C(t)=V₀e^(–t/RC). I(t)=–(Q₀/RC)e^(–t/RC) (negative=discharging direction). At t=τ: Q=Q₀/e≈0.368Q₀. At t=5τ: Q≈0. Heat dissipated in R during complete discharge: total=Q₀²/2C=initial energy stored. During charging: energy from battery=Cε²; energy in capacitor=½Cε²; heat=½Cε² (always lost regardless of R). JEE Main Current Electricity: τ=RC questions are direct substitution; "at what time is charge 50% of max" → 0.5Q₀=Q₀(1–e^(–t/RC)) → t=RC ln2=0.693RC.

In Current Electricity, cell combinations from Aakash PDF: n identical cells (EMF ε, internal resistance r) in SERIES: ε_eff=nε; r_eff=nr. Current through external R: I=nε/(R+nr). Preferred when R>>r (external resistance much larger than nr → internal drop small → nearly all EMF available). Example: flashlight batteries in series for more voltage. n identical cells in PARALLEL: ε_eff=ε; r_eff=r/n. Current: I=ε/(R+r/n). Preferred when R<ε₂: current from ε₁ through ε₂ may charge the weaker cell. JEE Main Current Electricity: n=4 identical cells each ε=2V, r=0.5Ω. Find I through R=1Ω for (a) all series → I=8/3≈2.67A; (b) all parallel → I=2/1.125≈1.78A. Choose series here since R=1Ω>r=0.5Ω.

In Current Electricity, power P=VI=I²R=V²/R. Three forms: given V and I directly → P=VI; given current and R → P=I²R; given voltage and R → P=V²/R. Commercial unit 1kWh=3.6×10⁶J. Household appliance rated "P watts at V volts" means: at supply voltage V, it consumes P watts. Resistance R=V²/P; full load current I=P/V. SERIES connection (same current I): P₁/P₂=R₁/R₂. Higher resistance → more power dissipation. Bulbs in series: brighter bulb has higher resistance. But note: rated power at rated voltage → actual R=V_rated²/P_rated. If bulbs have same rated voltage: R₁/R₂=V²/P₁ ÷ V²/P₂=P₂/P₁. Lower rated power → higher R → more power in series. PARALLEL connection (same voltage V): P=V²/R. Lower resistance → more power. Bulbs in parallel: brighter bulb has lower resistance (higher rated power at same voltage). Joule's heating: H=I²Rt=Pt. Heat in parallel combinations with identical resistors: H ∝ 1/R (if same V). JEE Main Current Electricity: "60W and 100W bulbs in series → which is brighter?" → series: same I → P=I²R → higher R → higher P. R₆₀>R₁₀₀ (since P_rated=V²/R → lower P means higher R) → 60W bulb is brighter in series.

In Current Electricity, meter bridge from Aakash PDF: a meter bridge uses a uniform resistance wire of length 100cm. It's a practical Wheatstone bridge. Setup: unknown X in left gap; known resistance R in right gap. Wire connected between them. Galvanometer connects wire to junction of X and R. Jockey moved along wire until galvanometer shows zero (null point at length l from left). From Wheatstone balance condition (P=resistance of wire up to l; Q=rest of wire; X and R are the other two arms): X/P=R/Q → X/l=R/(100–l) → X=Rl/(100–l). Alternatively: X/R=l/(100–l). Accuracy: most accurate when l≈50cm (balance near middle). Sources of error: non-uniform wire, contact resistance at terminals, end corrections. End correction: l is replaced by (l+e₁) and (100–l+e₂). To verify: interchange X and R → new balance at l'. Check if l×(100–l')=(100–l)×l' (cross-multiply). Sensitivity: galvanometer sensitivity determines minimum detectable change in l. To measure X accurately: R should be chosen so that l is neither too small (<20) nor too large (>80). JEE Main Current Electricity: "balance at l=60cm with R=10Ω" → X=10×60/40=15Ω.

In Current Electricity, temperature effect from Aakash PDF: R=R₀(1+αΔT) where α=temperature coefficient of resistance (per °C or per K). Three categories: (1) Metals (conductors): positive α (typically α≈4×10⁻³/°C for copper, 3.9×10⁻³/°C for tungsten). Resistance increases with T because τ decreases (more lattice vibrations → more collisions → shorter relaxation time → higher ρ). (2) Semiconductors (Si, Ge): negative α. Resistance decreases with T because n (charge carrier density) increases rapidly with T (more electron-hole pairs excited thermally) → conductivity σ=ne²τ/m increases despite τ decreasing. (3) Alloys (manganin, nichrome, constantan): α≈0 (very small). Used in standard resistors and heaters (resistance nearly independent of T). Special cases: superconductors: ρ→0 below critical temperature T_c (zero resistance). Thermistor (NTC): large negative α → used in temperature sensing. PTC thermistor: positive α. Carbon resistors: slightly negative α. JEE Main Current Electricity: "resistance of wire at 100°C is 10Ω. Find at 0°C if α=0.004/°C" → R₁₀₀=R₀(1+α×100) → 10=R₀×1.4 → R₀=10/1.4≈7.14Ω.