Physics Current Electricity

Current Electricity Formula Sheet

All key Current Electricity formulas for JEE Physics: current, Ohm's law, resistance, power, Kirchhoff's laws, Wheatstone & meter bridge. Quick revision for JEE Main & Advanced.

7 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know formula and result from the Current Electricity chapter, grouped by sub-topic for last-minute revision. Use the tables to scan fast and the boxed formulas for the headline relations.

Electric Current and Drift Velocity

$$\boxed{I = \frac{dQ}{dt} = \frac{Q}{t}}$$$$\boxed{v_d = \frac{eE\tau}{m}} \qquad \boxed{I = nAev_d} \qquad v_d = \frac{I}{nAe}$$
QuantityFormulaNotes
Electric current$I = \dfrac{Q}{t}$Unit: ampere (A) = C/s
Drift velocity$v_d = \dfrac{eE\tau}{m}$$\tau$ = relaxation time
Drift velocity (from $I$)$v_d = \dfrac{I}{nAe}$Most useful JEE form
Current from drift$I = nAev_d$Links micro to macro
Current density$J = \dfrac{I}{A} = nev_d$Unit: A/m²
Current density (vector)$\vec{J} = ne\vec{v_d}$Direction opposite to electron drift
Mobility$\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m}$Unit: m²/(V·s)
Conductivity–mobility$\sigma = ne\mu$
Current via mobility$I = neA\mu E = neA\mu\dfrac{V}{L}$
Drift velocity (same V)$v_d = \dfrac{V}{\rho L ne}$$v_d \propto 1/L$ for same V

Microscopic Ohm’s law:

$$\boxed{\vec{J} = \sigma \vec{E}} \qquad \sigma = \frac{ne^2\tau}{m}$$
High-Yield Reminders

Stretched wire (same current): $v_d \propto 1/A$, so if length doubles ($A$ halves), $v_d$ doubles.

Drift velocity is tiny ($\sim 10^{-4}$ m/s) but the signal speed $\approx c = 3\times10^8$ m/s — that is why the bulb lights instantly.

Copper number density: $n \approx 8.5 \times 10^{28}$ electrons/m³ (about one free electron per atom).

Ohm’s Law and V–I Characteristics

$$\boxed{V = IR}$$
FormExpressionUse when
Current$I = \dfrac{V}{R}$$V$, $R$ known
Resistance$R = \dfrac{V}{I}$from V–I measurement
Microscopic$J = \sigma E$vector / material form

Units: Voltage = volt (J/C), Current = ampere (C/s), Resistance = ohm (V/A).

Material typeV–I graphObeys Ohm’s law?
Metallic conductor (const. T)Straight line through originYes
Filament bulbCurves (R rises with T)No
Diode / LEDThreshold then steep riseNo
ElectrolyteNon-linear at low VNo (initially)
Graph & Slope Traps

V–I graph (V on y-axis): slope $= R$. I–V graph (I on y-axis): slope $= 1/R$ = conductance.

For a curved (non-ohmic) graph, $R = V/I$ at that point; dynamic resistance $= dV/dI$.

Ohm’s law is not universal — it holds only for ohmic conductors at constant temperature.

Resistance, Resistivity and Temperature

$$\boxed{R = \rho\frac{L}{A}} \qquad \boxed{R_T = R_0(1 + \alpha\Delta T)}$$
QuantityFormulaNotes
Resistance$R = \rho\dfrac{L}{A}$$R \propto L$, $R \propto 1/A$
Resistivity$\rho = R\dfrac{A}{L}$Material property (Ω·m)
Conductivity$\sigma = \dfrac{1}{\rho}$Unit: S/m
Conductance$G = \dfrac{1}{R} = \dfrac{I}{V} = \dfrac{\sigma A}{L}$Unit: siemens (S)
Temperature (metal)$R_T = R_0[1+\alpha(T-T_0)]$$\alpha > 0$ for metals
Temperature (semicond.)$R_T = R_0 e^{-\beta\Delta T} \approx R_0(1-\alpha\Delta T)$$\alpha < 0$
Compare two wires$\dfrac{R_1}{R_2} = \dfrac{\rho_1}{\rho_2}\cdot\dfrac{L_1}{L_2}\cdot\dfrac{A_2}{A_1}$ratio method
Hollow cylinder$R = \dfrac{\rho L}{\pi(r_2^2 - r_1^2)}$$r_2$ outer, $r_1$ inner

Increasing resistivity order: Silver < Copper < Gold < Aluminium (in $\rho$); copper $\rho \approx 1.7\times10^{-8}$ Ω·m.

Material$\alpha$ (per °C)Behaviour
Metals (Cu, Ag, Al, W)$\sim +4\times10^{-3}$R increases with T
Nichrome, Manganin$\sim 10^{-4}$ or lessnearly constant
Semiconductors, electrolytesNegativeR decreases with T
Wire Stretching & Cutting (JEE Favourites)

Stretched to $n\times$ length (volume constant): $\boxed{R' = n^2 R}$.

Radius halved: length $\times 4$, area $\div 4$ ⟹ $R' = 16R$.

Cut into $n$ equal parts, all in parallel: $\boxed{R_{eq} = \dfrac{R}{n^2}}$.

Combination of Resistors

$$\boxed{R_{eq} = R_1 + R_2 + R_3 + \dots} \quad \text{(series)}$$$$\boxed{\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots} \quad \text{(parallel)}$$
ConfigurationResultNotes
Two in parallel$R_{eq} = \dfrac{R_1 R_2}{R_1+R_2}$“product over sum”
$n$ equal in series$R_{eq} = nR$same current in all
$n$ equal in parallel$R_{eq} = \dfrac{R}{n}$same voltage across all
Voltage divider (series)$V_1 = \dfrac{R_1}{R_1+R_2}V$voltage $\propto R$
Current divider (parallel)$I_1 = \dfrac{R_2}{R_1+R_2}I$current $\propto 1/R$
Conductance in parallel$G_{eq} = G_1 + G_2 + G_3$add directly

EMF and Internal Resistance

$$\boxed{V = \varepsilon - Ir} \qquad I = \frac{\varepsilon}{R+r}$$
ConfigurationEMFInternal resistance
Cells in series$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$$r_{eq} = r_1 + r_2$
Cells in parallel$\varepsilon_{eq} = \dfrac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2}$$\dfrac{1}{r_{eq}} = \dfrac{1}{r_1}+\dfrac{1}{r_2}$
$n$ identical cells parallel$\varepsilon_{eq} = \varepsilon$$r_{eq} = \dfrac{r}{n}$
Terminal Voltage

On discharge $V = \varepsilon - Ir$ (terminal V < EMF). Open circuit: $V = \varepsilon$. Short circuit: $I = \varepsilon/r$ (maximum current).

Electrical Power and Heating

$$\boxed{P = VI = I^2R = \frac{V^2}{R}}$$$$\boxed{H = I^2Rt = Pt}$$
QuantityFormulaUse when
Power (basic)$P = VI$$V$, $I$ given
Power (current form)$P = I^2R$series — same $I$
Power (voltage form)$P = \dfrac{V^2}{R}$parallel — same $V$
Heat (Joule’s law)$H = I^2Rt = VIt = \dfrac{V^2t}{R}$energy in joules
Heat in calories$H = \dfrac{I^2Rt}{4.2}$1 cal = 4.2 J
Energy$E = Pt = VIt = I^2Rt = \dfrac{V^2t}{R}$
Bulb resistance from rating$R = \dfrac{V^2}{P}$$V$, $P$ = rated values
Commercial unit$1\text{ kWh} = 3.6\times10^6$ J“1 unit” of electricity

Maximum power transfer: load $R$ = source internal resistance $r$.

$$\boxed{R = r} \qquad P_{max} = \frac{\varepsilon^2}{4r} \qquad \eta = 50\%$$
Bulb Brightness — Series vs Parallel

Series (same $I$, use $P = I^2R$): higher R glows brighter (a 60 W bulb beats a 100 W bulb).

Parallel (same $V$, use $P = V^2/R$): lower R / higher rating glows brighter (100 W beats 60 W).

A “100 W, 220 V” rating holds only at rated voltage; at any other $V$ use $P = V^2/R$ with $R = V_{rated}^2/P_{rated}$ constant.

Heater coils — time $\propto R$ for fixed heat: series adds times ($t_1+t_2$), parallel adds reciprocals ($1/t = 1/t_1 + 1/t_2$).

Kirchhoff’s Laws

$$\boxed{\sum I_{in} = \sum I_{out}} \quad \text{(KCL — charge conservation)}$$$$\boxed{\sum V = 0} \quad \text{(KVL — energy conservation)}$$

KVL alternative form: $\sum \varepsilon = \sum IR$.

Element / traversalPotential changeNotes
Resistor, with current$-IR$ (drop)current flows high → low V
Resistor, against current$+IR$ (rise)
Battery, $-$ to $+$$+\varepsilon$ (rise)“uphill”
Battery, $+$ to $-$$-\varepsilon$ (drop)“downhill”
Number of KCL equations$(n-1)$ for $n$ junctionsnth is redundant
Number of KVL equationsone per independent loopuse mesh loops
Sign Convention — the #1 JEE Error

Traversing a resistor with the current is a potential drop ($-IR$); against it is a rise ($+IR$).

If a solved current comes out negative, the actual direction is opposite to the one you assumed.

Check: $\sum P_{\text{sources}} = \sum P_{\text{loads}}$ (power balance).

Wheatstone Bridge and Meter Bridge

$$\boxed{\frac{P}{Q} = \frac{R}{S}} \quad\Longleftrightarrow\quad \boxed{P\times S = Q\times R}$$

At balance, no current flows through the galvanometer and $V_C = V_D$.

QuantityFormulaNotes
Balance condition$\dfrac{P}{Q} = \dfrac{R}{S}$, i.e. $PS = QR$products of opposite arms equal
Unknown resistance$R = \dfrac{P\times S}{Q}$three arms known
Balanced bridge $R_{eq}$$R_{eq} = \dfrac{(P+Q)(R+S)}{P+Q+R+S}$only when balanced
Meter bridge ($l$ from R side)$R = P\times\dfrac{100-l}{l}$$l$ in cm on 100 cm wire
Meter bridge ($l$ from P side)$R = P\times\dfrac{l}{100-l}$check which side $l$ is
Interchange gaps$l_2 = 100 - l_1$complementary balance point
End corrections$R = P\times\dfrac{100-l+\alpha}{l+\beta}$$\alpha,\beta$ end resistances
Shunt effect$R' = \dfrac{R\,R_{shunt}}{R+R_{shunt}}$effective R decreases
graph LR
    A((A)) -- P --> C((C))
    A -- R --> D((D))
    C -- Q --> B((B))
    D -- S --> B
    C -- "G (galvanometer)" --- D
    A -- "battery" --- B
Meter Bridge Quick Wins

If $P = R$, balance is at the centre, $l = 50$ cm. Best sensitivity is near $l = 50$ cm; poor near the ends.

Meter bridge wire is uniform, so resistance $\propto$ length — that is why the length ratio replaces the resistance ratio.

When balanced, current still flows through the P–Q and R–S branches; only the galvanometer branch carries none.

Galvanometer Conversion

$$\boxed{S = \frac{I_g G}{I - I_g}} \quad \text{(ammeter: shunt in parallel)}$$$$\boxed{R = \frac{V}{I_g} - G} \quad \text{(voltmeter: high R in series)}$$
ConversionAdded resistancePositionEffect on circuit
To ammeterlow shunt $S = \dfrac{I_g G}{I - I_g}$parallellow net resistance; connect in series
To voltmeterhigh $R = \dfrac{V}{I_g} - G$serieshigh net resistance; connect in parallel

Here $G$ = galvanometer resistance, $I_g$ = full-scale deflection current.

Potentiometer

Principle: potential drop $\propto$ length, so $\dfrac{V_1}{V_2} = \dfrac{l_1}{l_2}$.

ApplicationFormulaNotes
Comparing EMFs$\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$balancing lengths
Internal resistance$r = R\left(\dfrac{l_1 - l_2}{l_2}\right)$$l_1$ open, $l_2$ with $R$ across cell
Why a Potentiometer Beats a Voltmeter

At balance it draws no current from the cell, so it measures true EMF — more accurate than a voltmeter, which always draws some current.

Constants and Standard Values

ConstantValue
Electron charge $e$$1.6 \times 10^{-19}$ C
Electron mass $m$$9.1 \times 10^{-31}$ kg
Number density (copper)$\approx 8.5 \times 10^{28}$ m⁻³
Resistivity of copper$\approx 1.7 \times 10^{-8}$ Ω·m
1 kWh$3.6 \times 10^{6}$ J
1 calorie$4.2$ J