Current Electricity Formula For JEE 2026
Current Electricity is one of the most important and high-scoring chapters in JEE Physics, with around 2–3 questions appearing almost every year in the exam. This chapter covers key concepts such as Ohm’s Law, drift velocity, resistance and resistivity, Kirchhoff’s laws, Wheatstone bridge, potentiometer, EMF, internal resistance, and electric power. A clear understanding of these concepts helps students solve circuit-based numerical problems quickly and accurately in the exam.
Many questions in JEE Mains and JEE Advanced are directly based on the formulas from this chapter, which makes regular revision very important for aspirants. To make revision easier and faster, students often use a JEE Physics Formula PDF that compiles all the important formulas in one place, helping them quickly review key concepts during practice sessions and last-minute preparation.
Electric Current and Drift Velocity Formulas
In the previous topic (Electrostatics), we studied charges at rest. Now we study charges in motion. When charges flow through a conductor (like a metal wire), we get an electric current. Think of it like water flowing through a pipe — the "flow rate" of charge is what we call current.
Electric Current: The rate of flow of electric charge through a cross-section of a conductor: $$I = Q/t$$, or more precisely $$I = dQ/dt$$. SI unit: ampere (A).
Ampere: One ampere means one coulomb of charge flowing per second: $$1 \; \text{A} = 1 \; \text{C/s}$$.
By convention, the direction of current is the direction in which positive charges would move. In a metal wire, electrons (negative charges) actually move, so the conventional current direction is opposite to the electron flow.
Types of Current
- Direct Current (DC): Current flows in one direction only. Example: battery.
- Alternating Current (AC): Current reverses direction periodically. Example: household electricity.
- Steady Current: The magnitude and direction of current do not change with time.
Drift Velocity
Inside a metal wire, free electrons move randomly in all directions at very high speeds (~$$10^5$$ m/s). Without an electric field, their average velocity is zero. When a battery is connected, the electric field pushes electrons slowly in one direction. This slow, net movement is called drift velocity.
Drift Velocity ($$v_d$$): The average velocity with which free electrons drift through a conductor under the influence of an electric field. It is very small — typically ~$$10^{-4}$$ m/s.
Drift Velocity and Current
$$$I = nAv_d e$$$
where:
- $$n$$ = number density of free electrons (electrons per unit volume, in m$$^{-3}$$)
- $$A$$ = cross-sectional area of the conductor
- $$v_d$$ = drift velocity
- $$e$$ = charge of an electron ($$1.6 \times 10^{-19}$$ C)
Also: $$v_d = \dfrac{eE\tau}{m}$$, where $$E$$ is the electric field, $$\tau$$ is the relaxation time, and $$m$$ is the electron mass.
Current Density ($$J$$): Current per unit cross-sectional area: $$J = I/A = nev_d$$. SI unit: A/m$$^2$$. In vector form: $$\vec{J} = ne\vec{v}_d$$.
Worked Example
A copper wire of cross-section $$1 \; \text{mm}^2$$ carries a current of 1.1 A. If copper has $$8.5 \times 10^{28}$$ free electrons per m$$^3$$, find the drift velocity.
$$v_d = \dfrac{I}{nAe} = \dfrac{1.1}{8.5 \times 10^{28} \times 1 \times 10^{-6} \times 1.6 \times 10^{-19}}$$
$$= \dfrac{1.1}{8.5 \times 10^{28} \times 1.6 \times 10^{-25}} = \dfrac{1.1}{13600} = 8.1 \times 10^{-5}$$ m/s $$\approx 0.08$$ mm/s
Notice how incredibly slow the drift velocity is!
Tip: Even though drift velocity is tiny, the electric signal (field) travels at nearly the speed of light. That is why a bulb turns on instantly when you flip a switch — you do not have to wait for electrons to travel from the switch to the bulb.
Ohm's Law
In 1827, Georg Simon Ohm discovered that for many materials, the current through them is directly proportional to the voltage applied across them (at constant temperature). This simple relationship is called Ohm's law.
Ohm's Law
$$$V = IR$$$
where:
- $$V$$ = potential difference across the conductor (in volts)
- $$I$$ = current through the conductor (in amperes)
- $$R$$ = resistance of the conductor (in ohms, $$\Omega$$)
Materials that obey Ohm's law are called ohmic (e.g., metals at constant temperature).
Materials that do not obey it are called non-ohmic (e.g., diodes, transistors).
Resistance and Resistivity Formulas
Resistance is a measure of how much a conductor opposes the flow of current. A thicker, shorter wire has less resistance; a thinner, longer wire has more resistance just like water flows more easily through a short, wide pipe.
Resistance ($$R$$): The ratio $$V/I$$ for a conductor. SI unit: ohm ($$\Omega$$). $$1\;\Omega = 1$$ V/A.
Resistance of a Wire
$$$R = \rho \frac{l}{A}$$$
where:
- $$\rho$$ = resistivity of the material ($$\Omega \cdot$$m)
- $$l$$ = length of the wire
- $$A$$ = cross-sectional area
Resistivity ($$\rho$$): A material property that quantifies how strongly the material resists current flow. Low $$\rho$$ = good conductor (metals). High $$\rho$$ = insulator.
Conductance: The reciprocal of resistance: $$G = 1/R$$. SI unit: siemens (S) or $$\Omega^{-1}$$.
Conductivity ($$\sigma$$): The reciprocal of resistivity: $$\sigma = 1/\rho$$. SI unit: S/m.
Temperature Dependence of Resistance
For metals, resistance increases with temperature because atoms vibrate more and scatter electrons more frequently. For semiconductors, resistance decreases with temperature because more charge carriers become available.
Temperature Dependence
$$$R_T = R_0(1 + \alpha \Delta T)$$$
$$$\rho_T = \rho_0(1 + \alpha \Delta T)$$$
where:
- $$R_0$$, $$\rho_0$$ = resistance and resistivity at reference temperature (usually 0°C or 20°C)
- $$\alpha$$ = temperature coefficient of resistance (unit: per °C or per K)
- $$\Delta T = T - T_0$$ = change in temperature
- For metals: $$\alpha > 0$$ (resistance increases with temperature)
- For semiconductors: $$\alpha < 0$$ (resistance decreases with temperature)
Worked Example
A copper wire has a resistance of 10 $$\Omega$$ at 20°C. If $$\alpha = 3.9 \times 10^{-3}$$ per °C, find its resistance at 100°C.
$$R_{100} = R_0(1 + \alpha \Delta T) = 10(1 + 3.9 \times 10^{-3} \times 80)$$
$$= 10(1 + 0.312) = 10 \times 1.312 = 13.12 \; \Omega$$
Colour Coding of Resistors
Resistors used in electronic circuits are often too small to have their values printed on them. Instead, coloured bands are painted on them. Each colour represents a digit or a multiplier.
Resistor Colour Code
| Colour | Digit | Multiplier |
|---|---|---|
| Black | 0 | $$10^0 = 1$$ |
| Brown | 1 | $$10^1$$ |
| Red | 2 | $$10^2$$ |
| Orange | 3 | $$10^3$$ |
| Yellow | 4 | $$10^4$$ |
| Green | 5 | $$10^5$$ |
| Blue | 6 | $$10^6$$ |
| Violet | 7 | $$10^7$$ |
| Grey | 8 | $$10^8$$ |
| White | 9 | $$10^9$$ |
| Gold | — | $$10^{-1}$$ (tolerance $$\pm 5\%$$) |
| Silver | — | $$10^{-2}$$ (tolerance $$\pm 10\%$$) |
For a 4-band resistor: 1st band (1st digit) + 2nd band (2nd digit) + 3rd band (multiplier) + 4th band (tolerance).
Tip: A popular mnemonic: B B R O Y of G B V G W ("BB ROY of Great Britain has a Very Good Wife") for the sequence Black, Brown, Red, Orange, Yellow, Green, Blue, Violet, Grey, White.
Worked Example
A resistor has colour bands: Yellow, Violet, Orange, Gold. Find its resistance.
Yellow = 4, Violet = 7, Orange = $$10^3$$, Gold = $$\pm 5\%$$
Resistance $$= 47 \times 10^3 \; \Omega = 47 \; \text{k}\Omega \; (\pm 5\%)$$
Series and Parallel Resistance Formulas
Resistors can be connected in two basic ways: series (end to end, same current flows through each) and parallel (side by side, same voltage across each).
Series and Parallel Combinations
Series (same current through all):
$$$R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots$$$
Parallel (same voltage across all):
$$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$$$
For two resistors in parallel: $$R_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2}$$
Worked Example
Three resistors of 2 $$\Omega$$, 3 $$\Omega$$, and 6 $$\Omega$$ are connected in parallel. Find the equivalent resistance.
$$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3 + 2 + 1}{6} = \dfrac{6}{6} = 1$$
$$R_{\text{eq}} = 1 \; \Omega$$
Kirchhoff's Laws
For circuits more complex than simple series/parallel, we use Kirchhoff's laws. These are based on two fundamental conservation principles: conservation of charge and conservation of energy.
Kirchhoff's Junction Rule (KCL)
At any junction (node) in a circuit, the total current entering equals the total current leaving:
$$$\sum I_{\text{in}} = \sum I_{\text{out}}$$$
This is based on conservation of charge — charge cannot pile up at a point.
Kirchhoff's Loop Rule (KVL)
In any closed loop of a circuit, the algebraic sum of all potential differences is zero:
$$$\sum \Delta V = 0$$$
This is based on conservation of energy — a charge returning to its starting point has the same potential.
Sign conventions for traversing a loop:
- Across a resistor in the direction of current: $$\Delta V = -IR$$ (voltage drop)
- Across a resistor against the direction of current: $$\Delta V = +IR$$
- Across a battery from $$-$$ to $$+$$ terminal: $$\Delta V = +\mathcal{E}$$
- Across a battery from $$+$$ to $$-$$ terminal: $$\Delta V = -\mathcal{E}$$
Worked Example
In a loop, a 12 V battery drives current $$I$$ through resistors of 2 $$\Omega$$ and 4 $$\Omega$$ in series. Find $$I$$.
Applying KVL around the loop:
$$+12 - I(2) - I(4) = 0$$
$$12 = 6I$$
$$I = 2$$ A
Wheatstone Bridge
The Wheatstone bridge is a clever circuit arrangement used to measure unknown resistances with high precision. It consists of four resistors arranged in a diamond shape with a galvanometer connected across the middle.
Wheatstone Bridge — Balanced Condition
When the galvanometer shows zero deflection (no current through it), the bridge is balanced and:
$$$\frac{P}{Q} = \frac{R}{S}$$$
where $$P$$, $$Q$$, $$R$$, $$S$$ are the four resistances.
If three are known, the fourth can be calculated.
Tip: When a Wheatstone bridge is balanced, the galvanometer branch can be removed without affecting the rest of the circuit. This simplifies analysis greatly.
Meter Bridge
A meter bridge (also called a slide wire bridge) is a practical version of the Wheatstone bridge. It uses a 1-metre long uniform wire as two of the resistances ($$R$$ and $$S$$), replacing them with lengths of the wire.
Meter Bridge
If the balance point is at distance $$l$$ from one end:
$$$\frac{R}{S} = \frac{l}{100 - l}$$$
where $$R$$ is the known resistance, $$S$$ is the unknown, and $$l$$ is in cm (the wire is 100 cm long).
So: $$S = R \cdot \dfrac{100 - l}{l}$$
Worked Example
In a meter bridge experiment, a known resistance of 30 $$\Omega$$ is in the left gap and an unknown resistance in the right gap. The balance point is at 40 cm from the left end. Find the unknown resistance.
$$\dfrac{30}{S} = \dfrac{40}{100 - 40} = \dfrac{40}{60} = \dfrac{2}{3}$$
$$S = 30 \times \dfrac{3}{2} = 45 \; \Omega$$
Potentiometer Formulas
A potentiometer is a device used to measure the EMF of a cell or the potential difference across a component without drawing any current from it. This makes it more accurate than a voltmeter (which always draws some current).
Potentiometer Principle: When a constant current flows through a wire of uniform cross-section, the potential difference across any length of the wire is proportional to that length.
Potentiometer Formulas
- Comparing two EMFs: $$$\frac{\mathcal{E}_1}{\mathcal{E}_2} = \frac{l_1}{l_2}$$$ where $$l_1$$ and $$l_2$$ are the balancing lengths.
- Internal resistance of a cell: $$$r = R\left(\frac{l_1 - l_2}{l_2}\right)$$$ where $$l_1$$ = balancing length with the cell in open circuit, $$l_2$$ = balancing length with the cell connected to external resistance $$R$$.
Worked Example
In a potentiometer experiment, a standard cell of EMF 1.02 V balances at 51 cm. An unknown cell balances at 75 cm. Find the EMF of the unknown cell.
$$\dfrac{\mathcal{E}_{\text{unknown}}}{1.02} = \dfrac{75}{51}$$
$$\mathcal{E}_{\text{unknown}} = 1.02 \times \dfrac{75}{51} = 1.02 \times 1.471 = 1.50$$ V
EMF and Internal Resistance Formulas
A battery (or cell) is a device that converts chemical energy into electrical energy. It maintains a potential difference between its terminals by doing work on charges. However, the battery itself has some resistance, called internal resistance.
EMF ($$\mathcal{E}$$): The electromotive force is the work done per unit charge by the battery in moving charge from the negative to the positive terminal inside the cell. It is the maximum voltage a battery can provide (when no current flows). Unit: volt (V).
Internal Resistance ($$r$$): The resistance offered by the electrolyte and electrodes inside the cell to the flow of current.
Terminal Voltage of a Cell
When a current $$I$$ flows:
$$$V = \mathcal{E} - Ir \quad \text{(discharging)}$$$
$$$V = \mathcal{E} + Ir \quad \text{(charging)}$$$
- $$V$$ = terminal voltage (what you measure across the battery's terminals)
- $$\mathcal{E}$$ = EMF of the cell
- $$I$$ = current flowing
- $$r$$ = internal resistance
- When $$I = 0$$ (open circuit): $$V = \mathcal{E}$$
Grouping of Cells
Series Grouping
$$n$$ identical cells, each with EMF $$\mathcal{E}$$ and internal resistance $$r$$, connected in series:
$$$\mathcal{E}_{\text{total}} = n\mathcal{E}, \quad r_{\text{total}} = nr$$$
$$$I = \frac{n\mathcal{E}}{R + nr}$$$
Best when external resistance $$R$$ is much larger than total internal resistance ($$R \gg nr$$).
Parallel Grouping
$$m$$ identical cells in parallel:
$$$\mathcal{E}_{\text{total}} = \mathcal{E}, \quad r_{\text{total}} = \frac{r}{m}$$$
$$$I = \frac{\mathcal{E}}{R + r/m}$$$
Best when external resistance $$R$$ is much smaller than internal resistance ($$R \ll r/m$$).
Mixed Grouping
$$n$$ cells in series per row, $$m$$ such rows in parallel:
$$$I = \frac{n\mathcal{E}}{R + nr/m}$$$
Maximum current when $$R = nr/m$$ (matching condition).
Maximum power is transferred to the external resistance when $$R = r_{\text{total}}$$.
Worked Example
12 cells, each of EMF 1.5 V and internal resistance 0.5 $$\Omega$$, are arranged in 3 rows of 4 cells each (mixed grouping). Find the current through an external resistance of $$\dfrac{2}{3}\;\Omega$$.
$$n = 4$$ (series), $$m = 3$$ (parallel), $$\mathcal{E} = 1.5$$ V, $$r = 0.5 \; \Omega$$, $$R = \dfrac{2}{3} \; \Omega$$
$$I = \dfrac{n\mathcal{E}}{R + nr/m} = \dfrac{4 \times 1.5}{2/3 + (4 \times 0.5)/3} = \dfrac{6}{2/3 + 2/3} = \dfrac{6}{4/3} = 6 \times \dfrac{3}{4} = 4.5$$ A
Electric Power and Joule Heating Formulas
When current flows through a resistor, electrical energy is converted into heat. This is why a filament bulb glows, a toaster heats up, and wires sometimes get warm. The rate at which electrical energy is consumed is called electric power.
Electric Power: The rate at which electrical energy is dissipated or consumed: $$P = VI$$. SI unit: watt (W). $$1 \; \text{W} = 1 \; \text{V} \times 1 \; \text{A}$$.
Power Dissipated in a Resistor
$$$P = VI = I^2 R = \frac{V^2}{R}$$$
- All three forms are equivalent (use Ohm's law to convert between them)
- Use $$P = I^2R$$ when current is known
- Use $$P = V^2/R$$ when voltage is known
Joule's Law of Heating
The heat produced in a resistor in time $$t$$ is:
$$$H = I^2 R t = VIt = \frac{V^2}{R} t$$$
- $$H$$ is in joules (J)
- 1 calorie = 4.186 J
- 1 kilowatt-hour (kWh) = $$3.6 \times 10^6$$ J (the "unit" of electricity)
Worked Example
An electric heater rated 1000 W, 220 V is used for 2 hours. Find: (a) the current drawn, (b) the resistance, (c) the energy consumed in kWh, (d) the cost if 1 kWh = Rs. 5.
(a) $$I = \dfrac{P}{V} = \dfrac{1000}{220} = 4.55$$ A
(b) $$R = \dfrac{V^2}{P} = \dfrac{(220)^2}{1000} = \dfrac{48400}{1000} = 48.4 \; \Omega$$
(c) Energy $$= P \times t = 1000 \times 2 = 2000$$ Wh $$= 2$$ kWh
(d) Cost $$= 2 \times 5 =$$ Rs. 10
Tip: When resistors are in series (same current): the one with higher resistance dissipates more power ($$P = I^2R$$). When in parallel (same voltage): the one with lower resistance dissipates more power ($$P = V^2/R$$).
Note: The maximum power delivered to an external resistance $$R$$ by a cell of EMF $$\mathcal{E}$$ and internal resistance $$r$$ is: $$$P_{\max} = \frac{\mathcal{E}^2}{4r} \quad \text{when } R = r \text{ (maximum power transfer theorem)}$$$
Current Electricity Formula for JEE 2026: Conclusion
Current Electricity is a fundamental chapter in JEE Physics that focuses on the behavior of electric charge in motion. Understanding concepts such as Ohm’s law, drift velocity, resistance, resistivity, Kirchhoff’s laws, Wheatstone bridge, and electric power helps students analyze electrical circuits effectively. Since many questions in JEE Mains and JEE Advanced are directly based on formulas from this chapter, mastering these formulas is essential for scoring well in the exam.
For effective preparation, students should regularly revise the Current Electricity formula sheet for JEE and practice numerical problems from different circuit configurations. Using a well-organized JEE Physics formula PDF can make revision faster and help aspirants recall important relationships quickly during the exam.
