## Lakhmir Singh Class 10 Physics Solutions Chapter 1 - Electricity - Free PDF Download

## FAQs on Lakhmir Singh Physics Class 10 Solutions Chapter 1 - Electricity

**1. What is an electric charge? Define 1 Coulomb of charge from the Lakhmir Singh Physics Class 10.**

The physical property of matter due to which it experiences a force on placing it in an electromagnetic field is called electric charge. There are two types of electric charges, namely, positive and negative charges. The sub-atomic particles that carry positive charges are called protons, and those carrying negative charges are called electrons. A naturally occurring element is always electrically neutral since the total number of electrons in it is equal to the total number of protons in its nucleus.

The SI unit of electric charge is Coulomb. The amount of charge that flows through the unit cross-section of a conductor, when a current of 1 ampere flows through it in one second is called 1 Coulomb.

**2. Differentiate between conductors and insulators with reference to Lakhmir Singh Physics Class 10?**

The materials that allow electricity to flow through them without any resistance are called conductors. The materials that offer resistance to the flow of electricity through them are called insulators. For example, all metals are conductors of electricity, and most non-metals are insulators. Copper, Aluminium, Silver, Graphite, are the common examples of conductors of electricity. Air, Bakelite, Plastic, Polythene are common examples of insulators.

**3. What is Coulomb’s law as explained in Lakhmir Singh Physics Class 10?**

Coulomb’s law states that the electrostatic force of attraction or repulsion between any two point charges is directly proportional to the product of their magnitudes and is inversely proportional to the square of the distance between them.

The expression for Coulomb's law is as follows.

F = kq_{1}q_{2}/r_{2}

Where F is the electrostatic force,

q_{1} and q_{2 }are the point charges

r is the distance between the points charges

And, k is Coulomb’s constant.

**4. Which is the best website to download the Lakhmir Singh Physics Class 10 Solutions Chapter-1 Electricity?**

Vedantu is the best website to download the Lakhmir Singh Physics Class 10 Solutions Chapter-1 Electricity. Our in-house team of subject-matter experts has prepared these solutions to facilitate an easy learning experience for the students of Class 10. These solutions are available in PDF format on Vedantu and all students can download them for free.

**5. What is Ohm's Law as explained in Lakhmir Singh Physics Class 10?**

Basically, it is the relationship between the resistance of an electric circuit and the current through that circuit. There are a number of mathematical formulas that attempt to predict the current through a certain circuit, based on its resistance. For example, Ohm's Law is given as

R = I k B T

Where

R = resistive load (light bulb, cord, mouse, etc)

I = current through the load

B = resistive impedance (difference between the medium to wire resistance and nominal load resistance)

T = the time taken for the current to travel through the circuit.

For example, for a light bulb that has a resistance of 60 ohms, with a current of 600mA and a time of 0.7ms,

R = 600 / (600/0.7) = 55 ohms

I = 600 / (0.7)/0.7 = 5.0 ohms

B = 5 ohms

T = 0.7/5 = 5ms

So, given a light bulb with a resistor value of 55 ohms, the current through the bulb is 5.0 ohms. In simple words, Ohm's Law predicts that there will be a current of 5.0 ohms for a time of 5ms. Of course, it is a limited model. If the load is completely blocked, the resistance of the circuit will become zero and the current will stop flowing.

Since Ohm's Law does not allow the model to handle loads with less than ideal currents, the circuit must be dissipated. One way of doing this is by removing all power sources from the circuit. For example, the light bulb can be removed. When the light bulb is removed, the current through the circuit will be zero. So, since there will be no current, Ohm's Law will not apply. Hence, the current through the circuit will drop to zero, and the light bulb won't provide any current. The line inductors and capacitors will then dissipate the current as heat.