Complete Formula List for Class 10 Science Electricity for Quick Revision (CBSE 2027)

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Formula Sheet

Complete formula guide for Electricity — CBSE Class 10 Science. Each formula is explained with notation, when to use, common mistakes, and exam tips.

Formulas

1

Electric Charge & Current

Quantization of charge and rate of flow
$$Q = ne \quad \text{and} \quad I = \frac{Q}{t}$$
Note: Charge of one electron (e) = 1.6 × 10-19 C
$I$
Electric CurrentA (Ampere)
$Q$
Total ChargeC (Coulomb)
$t$
Times (Seconds)
$n$
Number of electronsCount
  • Use $Q = ne$ when asked to find the number of electrons constituting a certain amount of charge.
  • Use $I = Q/t$ when calculating the steady current flowing through a conductor over a given time interval.
Forgetting to convert time given in minutes or hours into seconds before calculating current.
CBSE frequently asks to “Define 1 Ampere”. Write it derived directly from the formula: 1 Ampere is the current when 1 Coulomb of charge flows for 1 second.
2

Potential Difference & Ohm’s Law

Work done per unit charge and voltage-current relationship
$$V = \frac{W}{Q} \quad \text{and} \quad V = IR$$
Assumes temperature remains constant for Ohm’s Law.
$V$
Potential DifferenceV (Volt)
$W$
Work DoneJ (Joule)
$R$
Resistance$\Omega$ (Ohm)
  • Use $V = W/Q$ to find the energy required to move a charge across two points.
  • Use $V = IR$ in any circuit diagram problem to find missing voltage, current, or resistance.
Mixing up milliAmperes (mA) with Amperes. Always convert mA to A (divide by 1000) before plugging into Ohm’s Law.
The slope of a V-I graph gives the Resistance ($R$). The steeper the slope (if V is on the Y-axis), the higher the resistance!
3

Resistance & Resistivity

Factors affecting the resistance of a conductor
$$R = \rho \frac{l}{A} \quad \text{where} \quad A = \pi r^2 = \frac{\pi d^2}{4}$$
Resistivity ($\rho$) depends ONLY on the material and temperature, not dimensions.
$\rho$
Resistivity$\Omega \cdot m$
$l$
Length of conductorm (Meter)
$A$
Area of cross-section$m^2$
$r$ / $d$
Radius / Diameterm (Meter)
  • When calculating how resistance changes if a wire is stretched, cut, or doubled back on itself.
  • To identify the material of a wire by calculating its resistivity.
Using diameter directly as radius, or forgetting to convert mm/cm to meters before squaring the radius to find the Area.
The Stretched Wire Shortcut: If a wire is stretched to ‘$n$’ times its original length, its new resistance becomes $R’ = n^2 R$.
4

Equivalent Resistance

Series and Parallel Combinations
$$R_s = R_1 + R_2 + \dots \quad \text{and} \quad \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$$
Shortcut for 2 parallel resistors: $R_p = \frac{R_1 R_2}{R_1 + R_2}$
$R_s$
Total Resistance (Series)$\Omega$ (Ohm)
$R_p$
Total Resistance (Parallel)$\Omega$ (Ohm)
  • Use Series formula when resistors are connected end-to-end (Current is constant).
  • Use Parallel formula when resistors are connected across the same two points (Voltage is constant).
Calculating $\frac{1}{R_p}$ completely, but forgetting to take the reciprocal at the very end to find actual $R_p$.
For ‘$n$’ identical resistors of resistance $R$: Total in series is $nR$. Total in parallel is $\frac{R}{n}$.
5

Power & Joule’s Heating

Energy dissipation and commercial consumption
$$P = VI = I^2R = \frac{V^2}{R} \quad \text{and} \quad H = I^2Rt$$
Commercial Energy: $E = P \times t$ (1 kWh = $3.6 \times 10^6$ Joules)
$P$
Electric PowerW (Watt)
$H$
Heat EnergyJ (Joule)
$E$
Commercial EnergykWh
  • Use $P = I^2R$ or $H = I^2Rt$ for elements connected in Series (since I is constant).
  • Use $P = V^2/R$ or $H = (V^2/R)t$ for elements connected in Parallel (since V is constant).
  • Use $E = P \times t$ when calculating electricity bills.
For commercial energy ($E = P \times t$), failing to convert Power to Kilowatts (kW) and time to Hours (h).
If two bulbs are connected in series, the bulb with the lower power rating will glow brighter! If connected in parallel, the one with the higher power rating glows brighter.

Quick Reference

Formula Use When Missing Variable
$P = I^2R$ Comparing power dissipated in Series circuits V
$P = \frac{V^2}{R}$ Comparing power dissipated in Parallel circuits I
$R_p = \frac{R_1 R_2}{R_1 + R_2}$ Solving exactly 2 resistors in parallel quickly LCM / Fractions
$R = \rho \frac{l}{\pi r^2}$ Given the radius of a wire rather than cross-sectional area A

Concept Enhancement

★ Did You Know
The filament of an electric bulb is made of Tungsten because it has a very high melting point ($3380^\circ \text{C}$) and high resistivity, allowing it to get white-hot without melting.
⚡ Ammeter vs. Voltmeter Placement
An Ammeter is always connected in series because it needs to measure the total flow of current and has very low resistance. A Voltmeter is always connected in parallel across the points where potential difference is measured, as it has very high resistance to prevent drawing current from the main circuit.

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