Current Electricity formulas

Class 12 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 3, the special cases questions are built on, and diagrams where they help.

85 formulas14 sectionsClass 12 · Chapter 3Free sample · complete chapter

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Electric current

$$I=\frac{dq}{dt},\qquad q=\int_{t_1}^{t_2} I\,dt$$

Charge = area under the I–t graph. Average current $=\Delta q/\Delta t$. Unit: ampere (C s⁻¹).

$$I=\frac{nq}{t},\qquad I=\frac{(n_1+n_2)e}{t}$$

$n$ charges crossing in time $t$. Second form: $n_1$ positive ions one way and $n_2$ electrons the other way (discharge tube) – both add to the current.

$$I=qf=\frac{q\omega}{2\pi}=\frac{qv}{2\pi r}$$

Charge $q$ going round a circle (electron in an orbit, charged ring rotating).

Current has direction but is a scalar; conventional current is opposite to electron flow.

2Drift velocity, current density, mobility

$$v_d=\frac{eE\tau}{m},\qquad t_{\text{cross}}=\frac{l}{v_d}$$

$\tau$ = relaxation time ($\sim10^{-14}$ s); $v_d\sim10^{-4}$ m s⁻¹, opposite to $\vec E$ for electrons. Time for an electron to drift through a wire of length $l$.

$$I=neAv_d$$

Same current through a wire of varying area: $v_d\propto 1/A$ (faster in the thinner part).

$$J=\frac IA=nev_d,\qquad I=\int\vec J\cdot d\vec A=JA\cos\theta$$

$$\begin{array}{l}\displaystyle J=\frac IA=nev_d\\[6pt]\displaystyle I=\int\vec J\cdot d\vec A=JA\cos\theta\end{array}$$

Current density: a vector (A m⁻²); current is its flux.

$$\mu=\frac{v_d}{E}=\frac{e\tau}{m}$$

Mobility (m² V⁻¹ s⁻¹). Doubling $E$ doubles $v_d$ but not $\mu$.

3Ohm's law, resistance, resistivity

$$V=IR,\qquad R=\frac{\rho\,l}{A}$$

$R$ depends on size, material, temperature; $\rho$ only on material and temperature. Conductance $G=1/R$ (S).

$$R=\frac{ml}{ne^2\tau A},\quad \rho=\frac{m}{ne^2\tau},\quad \sigma=\frac1\rho=ne\mu$$

$$\begin{array}{l}\displaystyle R=\frac{ml}{ne^2\tau A}\\[6pt]\displaystyle \rho=\frac{m}{ne^2\tau}\\[6pt]\displaystyle \sigma=\frac1\rho=ne\mu\end{array}$$

Resistance from the electron picture.

$$\vec J=\sigma\vec E,\qquad E=\frac Vl$$

Ohm's law in vector (microscopic) form.

Non-ohmic: diodes, thermistors, gases, electrolytes, GaAs (curved, one-way or double-valued V–I graphs).
Slope of I–V graph $=1/R$; V–I graph $=R$. Steeper I–V line = lower $R$ (lower temperature for a metal).
$\rho_{\text{alloy}}>\rho_{\text{metals}}$, $\alpha_{\text{alloy}}<\alpha_{\text{metals}}$.

4Stretching and reshaping a wire

$$R=\frac{\rho l^2}{V}=\frac{\rho V}{A^2}\ \Rightarrow\ R\propto l^2\propto\frac1{A^2}\propto\frac1{r^4}$$

Volume fixed when a wire is stretched or drawn. Length ×$n$ → $R\times n^2$. Radius ÷$n$ → $R\times n^4$.

$$\frac{\Delta R}{R}=2\frac{\Delta l}{l}=-2\frac{\Delta A}{A}=-4\frac{\Delta r}{r}$$

Small stretch: length up $x\%$ → $R$ up $2x\%$; radius down $x\%$ → $R$ up $4x\%$. For a large $x\%$ stretch, $R$ rises by $\left(2x+\dfrac{x^2}{100}\right)\%$.

$$\frac{\Delta R}{R}=\frac{\Delta l}{l}\ (A\text{ fixed});\quad \frac{\Delta R}{R}=-2\frac{\Delta r}{r}\ (l\text{ fixed})$$

$$\begin{array}{l}\displaystyle \frac{\Delta R}{R}=\frac{\Delta l}{l}\ (A\text{ fixed});\\[6pt]\displaystyle \frac{\Delta R}{R}=-2\frac{\Delta r}{r}\ (l\text{ fixed})\end{array}$$

When only one dimension changes (e.g. two separate wires compared).

$$\text{Block } l>b>h:\ \ R_{\max}=\frac{\rho l}{bh},\quad R_{\min}=\frac{\rho h}{lb}$$

$$\begin{array}{l}\displaystyle \text{Block } l>b>h:\ \ R_{\max}=\frac{\rho l}{bh}\\[6pt]\displaystyle R_{\min}=\frac{\rho h}{lb}\end{array}$$

Use the length along the current and the area across it.

5Non-uniform conductors JEE Adv

$$R=\int\frac{\rho\,dx}{A(x)}$$

Slice the conductor along the current: series slices add as $\int dR$.

$$\text{Tapered wire (radii }a\to b):\ R=\frac{\rho l}{\pi ab}$$

Radius changing linearly along the length $l$.

$$\text{Radial, cylinder: }R=\frac{\rho}{2\pi l}\ln\frac ba,\qquad \text{sphere: }R=\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)$$

$$\begin{array}{l}\displaystyle \text{Radial, cylinder: }R=\frac{\rho}{2\pi l}\ln\frac ba\\[6pt]\displaystyle \text{sphere: }R=\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)\end{array}$$

Current flowing radially from the inner surface (radius $a$) to the outer ($b$).

$$\rho=\rho_0(1+\alpha x):\ \ R=\frac{\rho_0}{A}\left(L+\frac{\alpha L^2}{2}\right)$$

Resistivity varying along a uniform wire.

6Effect of temperature

$$R_t=R_0(1+\alpha t),\qquad \frac{R_1}{R_2}=\frac{1+\alpha t_1}{1+\alpha t_2}$$

$$\begin{array}{l}\displaystyle R_t=R_0(1+\alpha t)\\[6pt]\displaystyle \frac{R_1}{R_2}=\frac{1+\alpha t_1}{1+\alpha t_2}\end{array}$$

$\rho_2=\rho_1[1+\alpha(t_2-t_1)]$ for a small range. For large $t$: $R_t=R_0(1+\alpha t+\beta t^2)$.

$$\alpha=\frac{R_2-R_1}{R_1t_2-R_2t_1},\qquad \alpha=\frac1R\frac{dR}{dt}$$

From two readings (no $R_0$ given). If $t_1=0^\circ$C: $\alpha=\dfrac{R_2-R_1}{R_1t_2}$.

$$R_1\alpha_1+R_2\alpha_2=0$$

Two resistors in series with a total that does not change with temperature (one $\alpha$ must be negative).

Metals $\alpha>0$; semiconductors, insulators, electrolytes, ionised gases $\alpha<0$; alloys (manganin, constantan) tiny $\alpha$ → standard resistors, bridge wires.
Thermistor: large $|\alpha|$ (+ or −), used for small temperature changes. Superconductor: $\rho=0$ below $T_c$ (Hg 4.2 K).

7Colour code of carbon resistors NEET PYQ

Black 0	Brown 1	Red 2	Orange 3	Yellow 4
Green 5	Blue 6	Violet 7	Grey 8	White 9

B B R O Y of Great Britain had a Very Good Wife. Band 1, 2 = digits, band 3 = multiplier $10^n$, band 4 = tolerance: gold ±5% (multiplier $10^{-1}$), silver ±10% ($10^{-2}$), no band ±20%.

8Combination of resistors

$$R_s=R_1+R_2+\cdots\qquad \frac1{R_p}=\frac1{R_1}+\frac1{R_2}+\cdots$$

$$\begin{array}{l}\displaystyle R_s=R_1+R_2+\cdots\\[6pt]\displaystyle \frac1{R_p}=\frac1{R_1}+\frac1{R_2}+\cdots\end{array}$$

Series: same $I$, $V\propto R$. Parallel: same $V$, $I\propto 1/R$; $R_p$ is smaller than the smallest resistor.

$$V_1=\frac{R_1}{R_1+R_2}V,\qquad I_1=\frac{R_2}{R_1+R_2}I$$

Voltage divider (series); current divider (two in parallel).

$$R_{1,2}=\tfrac12\left(R_s\pm\sqrt{R_s^2-4R_sR_p}\right)$$

Finding two resistors from their series and parallel values ($R_sR_p=R_1R_2$, $R_s\ge4R_p$).

$$n\text{ equal: }\frac{R_s}{R_p}=n^2;\qquad \text{wire cut in }n,\ \text{all parallel: }\frac R{n^2}$$

$$\begin{array}{l}\displaystyle n\text{ equal: }\frac{R_s}{R_p}=n^2;\\[6pt]\displaystyle \text{wire cut in }n\\[6pt]\displaystyle \text{all parallel: }\frac R{n^2}\end{array}$$

Number of different values from $n$ equal resistors: $2^{\,n-1}$.

$$R_{\text{ring}}=\frac{R\,\theta(2\pi-\theta)}{4\pi^2}$$

Uniform ring of resistance $R$, between points $\theta$ apart (ends of a diameter: $R/4$).

$$R_{\text{polygon}}=\frac{(n-1)R}{n}$$

Closed polygon of $n$ sides, each $R$, across two adjacent corners. Wire stretched $m$ times then bent: use $m^2R$ for the total.

$$\text{Cube (12 edges, each }r):\ \ \frac{5r}{6},\ \ \frac{3r}{4},\ \ \frac{7r}{12}$$

$$\begin{array}{l}\displaystyle \text{Cube (12 edges, each }r):\ \ \frac{5r}{6}\\[6pt]\displaystyle \ \frac{3r}{4}\\[6pt]\displaystyle \ \frac{7r}{12}\end{array}$$

Across the body diagonal, a face diagonal and one edge. Found by symmetry – points at equal potential joined.

$$\text{Infinite ladder: } x=R_1+\frac{R_2\,x}{R_2+x}$$

Adding one more section does not change $x$. All $R$: $x=\dfrac{1+\sqrt5}{2}R$; series $2R$, rung $R$: $x=(1+\sqrt3)R$.

Look for the repeating section; the resistance to its right equals x again.

$$\text{Series wires: }\rho=\frac{\rho_1l_1+\rho_2l_2}{l_1+l_2};\qquad \text{parallel: }\rho=\frac{\rho_1\rho_2(A_1+A_2)}{\rho_1A_2+\rho_2A_1}$$

$$\begin{array}{l}\displaystyle \text{Series wires: }\rho=\frac{\rho_1l_1+\rho_2l_2}{l_1+l_2};\\[6pt]\displaystyle \text{parallel: }\rho=\frac{\rho_1\rho_2(A_1+A_2)}{\rho_1A_2+\rho_2A_1}\end{array}$$

Equivalent resistivity of two joined wires (equal areas / equal lengths).

How to simplify: name the nodes; points joined by a plain wire are one node; a resistor with both ends on one node carries no current.
Symmetry: points at equal potential can be joined or split; a resistor between them carries no current.
Balanced bridge inside a network: drop the middle resistor. Otherwise use Kirchhoff or the node method.

9Cell: emf, internal resistance, power

$$I=\frac{E}{R+r},\qquad V=E-Ir$$

Discharging ($VCharging: $V=E+Ir$. Open circuit $V=E$; short circuit $I=E/r$. Lost volts $=Ir$.

$$r=\left(\frac EV-1\right)R,\qquad r=\frac{I_2R_2-I_1R_1}{I_1-I_2}$$

$$\begin{array}{l}\displaystyle r=\left(\frac EV-1\right)R\\[6pt]\displaystyle r=\frac{I_2R_2-I_1R_1}{I_1-I_2}\end{array}$$

From terminal voltage, or from two readings ($I_1$ through $R_1$, $I_2$ through $R_2$).

$$\text{useful fraction}=\frac VE=\frac R{R+r},\qquad \text{lost}=\frac r{R+r}$$

$$\begin{array}{l}\displaystyle \text{useful fraction}=\frac VE=\frac R{R+r}\\[6pt]\displaystyle \text{lost}=\frac r{R+r}\end{array}$$

Voltmeter of resistance $R_V$ reads low by $\dfrac{r}{R_V+r}\times100\%$.

$$P_{\max}=\frac{E^2}{4r}\ \text{ at }R=r$$

Efficiency is 50% at maximum power. Same power in $R_1$ and in $R_2$ (separately) → $r=\sqrt{R_1R_2}$.

$r$ grows with electrode separation and electrolyte concentration, falls with electrode area. Emf depends only on the electrodes, electrolyte and temperature.

10Grouping of cells

$$\text{Series } (n\text{ cells}):\ I=\frac{nE}{R+nr}$$

$R\gg nr$: $I\approx nE/R$; $R\ll nr$: $I\approx E/r$ (series helps only when $R$ is large).

$$m\text{ of }n\text{ reversed: }\ E_{eq}=(n-2m)E,\ \ r_{eq}=nr$$

$$\begin{array}{l}\displaystyle m\text{ of }n\text{ reversed: }\ E_{eq}=(n-2m)E\\[6pt]\displaystyle \ r_{eq}=nr\end{array}$$

Two unlike cells in series: $E_1\pm E_2$, $r_1+r_2$; each cell's terminal voltage $V_k=E_k-Ir_k$ (can be zero!).

$$\text{Parallel } (n\text{ identical}):\ I=\frac{E}{R+r/n}=\frac{nE}{nR+r}$$

Parallel helps only when $R$ is small compared with $r/n$.

$$E_{eq}=\frac{E_1r_2+E_2r_1}{r_1+r_2},\qquad r_{eq}=\frac{r_1r_2}{r_1+r_2}$$

$$\begin{array}{l}\displaystyle E_{eq}=\frac{E_1r_2+E_2r_1}{r_1+r_2}\\[6pt]\displaystyle r_{eq}=\frac{r_1r_2}{r_1+r_2}\end{array}$$

Two unlike cells in parallel; a reversed cell enters with $-E$. General: $E_{eq}=\dfrac{\sum E_i/r_i}{\sum 1/r_i}$.

$$\text{Mixed: }I=\frac{mnE}{mR+nr}$$

$n$ cells in each row, $m$ rows in parallel. Maximum current when $R=\dfrac{nr}{m}$ (external = internal).

$$I_{\max}=\frac{nE}{2R}=\frac{mE}{2r},\qquad P_{\max}=mn\frac{E^2}{4r}$$

Values at that best grouping.

11Kirchhoff's laws

$$\sum I=0\ (\text{junction}),\qquad \sum IR+\sum E=0\ (\text{loop})$$

$$\begin{array}{l}\displaystyle \sum I=0\ (\text{junction})\\[6pt]\displaystyle \sum IR+\sum E=0\ (\text{loop})\end{array}$$

Junction rule: conservation of charge. Loop rule: conservation of energy.

Signs: across $R$ along the current $-IR$, against it $+IR$; across a cell from − to + $+E$, whatever the current.
Guess directions; a negative answer means the current flows the other way. Loops needed = number of independent meshes.
Capacitor in steady state: no current in its branch, so no drop across resistors in series with it; $Q=CV$ with $V$ between its ends.
Node method (fastest for JEE): one node at 0 V, write $\sum\frac{V-V_k}{R}=0$ at the others.

12Wheatstone bridge and metre bridge

$$\frac PQ=\frac RS\ \ (I_g=0),\qquad R_{eq}=\frac{(P+Q)(R+S)}{P+Q+R+S}$$

$$\begin{array}{l}\displaystyle \frac PQ=\frac RS\ \ (I_g=0)\\[6pt]\displaystyle R_{eq}=\frac{(P+Q)(R+S)}{P+Q+R+S}\end{array}$$

Balanced: remove or short the galvanometer arm; cell and galvanometer can be swapped. Most sensitive when $P\approx Q\approx R\approx S$. Capacitor bridge: $C_1/C_2=C_3/C_4$.

$$V_B-V_D=\frac{I\,(QR-PS)}{P+Q+R+S}$$

Unbalanced bridge: current in the galvanometer goes B → D if $QR>PS$.

$$\frac RS=\frac{l}{100-l},\qquad \frac RS=\frac{l+a}{100-l+b}$$

$$\begin{array}{l}\displaystyle \frac RS=\frac{l}{100-l}\\[6pt]\displaystyle \frac RS=\frac{l+a}{100-l+b}\end{array}$$

Metre bridge ($R$ in left gap, $l$ in cm); with end corrections $a$, $b$. Swap $R$ and $S$ → balance at $100-l$.

$R$ in left gap increased (or right gap decreased) → balance point moves right.
Changing the bridge-wire radius or the cell does not move the balance point. Best accuracy near $l=50$ cm.

13Potentiometer NEET PYQ

$$k=\frac VL=\frac{E_0}{r_0+R_w+R_h}\cdot\frac{R_w}{L},\qquad E=k\,l$$

$$\begin{array}{l}\displaystyle k=\frac VL=\frac{E_0}{r_0+R_w+R_h}\cdot\frac{R_w}{L}\\[6pt]\displaystyle E=k\,l\end{array}$$

Potential gradient (driver cell $E_0$, $r_0$; wire $R_w$, length $L$; rheostat $R_h$). Draws no current at balance. Smaller $k$ → more sensitive. $k=\dfrac{I\rho_s}{\pi r^2}$ for wire of resistivity $\rho_s$.

$$\frac{E_1}{E_2}=\frac{l_1}{l_2},\qquad \frac{E_1+E_2}{E_1-E_2}=\frac{l_1}{l_2}$$

Comparing emfs: directly, or with the cells assisting ($l_1$) and opposing ($l_2$).

$$r=R\left(\frac{l_1-l_2}{l_2}\right)$$

Internal resistance: $l_1$ on open circuit, $l_2$ with shunt $R$ across the cell.

$$\frac{R_2}{R_1}=\frac{l_2-l_1}{l_1},\qquad e=\frac{IR_w}{L}\,l$$

Comparing resistances ($l_1$ across $R_1$, $l_2$ across $R_1+R_2$). Thermo-emf: a large resistance in the primary makes $k$ tiny (μV range).

14Electrical power and heating

$$P=VI=I^2R=\frac{V^2}R,\qquad H=I^2Rt\ \ \left(=\tfrac{I^2Rt}{4.2}\text{ cal}\right)$$

$$\begin{array}{l}\displaystyle P=VI=I^2R=\frac{V^2}R\\[6pt]\displaystyle H=I^2Rt\ \ \left(=\tfrac{I^2Rt}{4.2}\text{ cal}\right)\end{array}$$

1 kWh $=3.6\times10^6$ J; units $=\dfrac{\sum(\text{watts}\times\text{hours})}{1000}$.

$$R=\frac{V_s^2}{P_s},\qquad P=P_s\left(\frac V{V_s}\right)^2$$

Appliance rated $(P_s,V_s)$ used on $V$. Same $V$ with $R\to R/n$: $P\to nP$.

$$\text{Series: }\frac1P=\sum\frac1{P_i};\qquad \text{parallel: }P=\sum P_i;\qquad P_p=n^2P_s$$

$$\begin{array}{l}\displaystyle \text{Series: }\frac1P=\sum\frac1{P_i};\\[6pt]\displaystyle \text{parallel: }P=\sum P_i;\\[6pt]\displaystyle P_p=n^2P_s\end{array}$$

Series: $P\propto R$ – lower-watt bulb glows brighter. Parallel: $P\propto 1/R$ – higher-watt bulb brighter. $n$ equal resistors: parallel gives $n^2$ times the series power.

$$t_s=t_1+t_2,\qquad t_p=\frac{t_1t_2}{t_1+t_2}$$

Time for two heater coils to give the same heat on the same supply ($t\propto R$).

$$\text{Fuse: } I^2\propto r^3,\ \text{independent of length};\quad t_{\text{melt}}\propto r^4$$

$$\begin{array}{l}\displaystyle \text{Fuse: } I^2\propto r^3\\[6pt]\displaystyle \text{independent of length};\\[6pt]\displaystyle t_{\text{melt}}\propto r^4\end{array}$$

Fuse wire: high $\rho$, low melting point (tin–lead). Transmission loss $=P^2R_c/V^2$ → transmit at high $V$.

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