# Class 10 Electricity Formulas

In this article we will try to provide all physics electricity formulas for class 10 science chapter 12.

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The movement of charge in a conductor from positive terminal to negative terminal is referred to as electricity. Electricity has a wide range of uses and applications. It serves as a medium for supplying power to electrical devices. We already know that the flow of charge generates current, which we call electricity.

To understand how electricity is produced, we must first understand the various basic parameters associated with it, such as voltage, current, resistance, conductivity, and the relationships between these quantities.

## Formulas of electricity

Given below are all the formulas used for class 10 electricity chapter.

1. Charge q on a body is always denoted by
$q = ne$
where n = any integer positive or negative and $e =1.602 \times 10^{-19} C$ i. e., charge on an electron or proton.
2. $Work\; done = charge \times potential$ or potential difference
Mathematically,
$W=qV=q(V_2V_1)\; Joule$
3. $Electric\; Current = \frac{charge}{time}$
Or,
$I=\frac{q}{t}\;Ampere$
4. Ohm’s Law
$Resistance=\frac{potential\; difference}{current}$
Or,
$R=\frac{V}{I}$
5. Resistance in terms of resistivity
$R=\frac{\rho l}{A}$
6. Resistance in series combination
$R_s=R_1+R_2+R_3+…..$
7. Resistance in parallel combination
$\frac{1}{R_p}= \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}+…..$
8. Electrical Power
$P=VI=I^2R=\frac{V^2}{R}$
9. Power in Series
$\frac{1}{P_s}= \frac{1}{P_1}+ \frac{1}{P_2}+ \frac{1}{P_3}+…..$
10. Power in parallel
$P_p=P_1+P_2+P_3+…..$
11. Electrical Energy
$E-Vit=I^2Rt=\frac{V^2}{R}t$
12. Heat produced
$H-Vit=I^2Rt=\frac{V^2}{R}t$

## Solved Examples

Question 1
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution
Here,
Battery Voltage $V= 12 V$
Current in circuit $I=2.5mA=2.5\times 10^{-3}A$
Resistance of the resistor $R=?$ is to be calculated
We know that from Ohm’s Law
$R=\frac{V}{I}$
Putting in the values we get
$R=\frac{12V}{2.5\times 10^{-3}A}$
$=4.8\times 10^{3}VA^{-1}=4.8\times 10^{3}\Omega$

Question 2
An electric heater of resistance $8\Omega$ draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution
It is given in the question that
Heater resistance $R=8\Omega$
Current drawn, $I=15 A$
Time for which current is drawn
$t=2h=2\times 60\times 60 \;sec=7200\; sec$
Rate of heat developed $P=?$ which is to be calculated
We know that
Rate of heat developed = rate at which power is consumed.
$P=I^2R$
Putting in the value we get,
$P=(15 A)^2\times 8\omega=1800W$
The rate of heat developed is
$P=1800 J/s$ Test Yourself on questions based on these formulas

## Test Yourself

General Instructions

• Your test contains multiple-choice questions with only one answer type of questions. There are a total of 5 questions
• This is a 10 min test. Please make sure you complete it in the stipulated time
• You can finish this test any time using the ‘View Results‘ button.

1. The law which gives a relation between electric potential and electric current is called
2. The correct relation between heat produced H and electric current I is
3. Which of the following terms does not represent electric power in a circuit
4. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’ then the ratio R/R’ is
5. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be

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op

thanks
Excellent understanding, best quality questions and there solutions . External support of brilliant understanding