# Cbse Science notes class-10th

## Chapter-12 Electricity

**Coulomb(C):**S.I unit of charge

**Current(I):**The rate of flow of charge is called current.

**Potential Difference(V):**work done to move a unit charge from one point to another.

**1 volt:when 1 joule work is done in carrying one coulomb charge then potential difference is called 1 volt.**

**1 Volt**:when 1 joule work is done in carrying one coulomb charge then potential difference is called 1 volt.

**Voltmeter:**Instrument to measure potential difference.

**Ohm's law:**Potential difference across the two points of a metallic conductor is directly proportional to current passing through the circuit provided that temperature remains constant.

**Resistance(R):**It is the property of a conductor to resist the flow of charges through it.

**Where potential difference is 1 V and current through the circuit is 1A , then resistance is 1Ohm.**

**Rheostat:**Variable resistance is a component used to regulate current without changing the source of voltage.

**NCERT Exercise Questions and Solutions:**

**Question 1:**

What does an electric circuit mean?

**Answer:**

An electric circuit is a closed path or loop through which electric current flows. It typically consists of electrical components like a source of electricity (battery or cell), a conductor (wires), and a load (bulb or resistor) that consume the energy.

**Question 2:**

Define the unit of current.

**Answer:**

The unit of electric current is the **ampere** (A). One ampere is defined as the flow of one coulomb of charge per second, i.e.,

$\text{1 A} = \text{1 C/s}$

**Question 3:**

Calculate the number of electrons constituting one coulomb of charge.

**Answer:**

The charge of one electron is $1.6 \times 10^{-19} \, \text{C}$.

Number of electrons in one coulomb, $n$, is given by:

$n = \frac{\text{Total charge}}{\text{Charge of one electron}} = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \text{ electrons}$**Question 4:**

Name a device that helps to maintain a potential difference across a conductor.

**Answer:**

A **cell** or a **battery** is a device that maintains a potential difference across a conductor.

**Question 5:**

What is meant by saying that the potential difference between two points is 1 V?

**Answer:**

A potential difference of 1 volt between two points means that 1 joule of work is required to move 1 coulomb of charge from one point to the other.

**Question 6:**

How much energy is given to each coulomb of charge passing through a 6 V battery?

**Answer:**

The energy $W$ given to each coulomb of charge is given by:

So, each coulomb of charge is given 6 joules of energy.

**Question 7:**

On what factors does the resistance of a conductor depend?

**Answer:**

The resistance of a conductor depends on:

**Length (L)**of the conductor - Resistance is directly proportional to length.**Cross-sectional area (A)**- Resistance is inversely proportional to the area.**Material**of the conductor - Different materials have different resistivities.**Temperature**- Higher temperature generally increases resistance.

**Question 8:**

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

**Answer:**

Current flows more easily through a **thick wire** because it has a larger cross-sectional area. According to the formula $R = \frac{\rho L}{A}$, a larger cross-sectional area $A$ results in a lower resistance, allowing more current to pass through.

**Question 9:**

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

**Answer:**

Alloys are used in the coils of electric toasters and irons because:

- They have higher resistivity, which produces more heat.
- They do not oxidise (burn) easily at high temperatures, ensuring durability.
- Alloys have higher melting points, making them suitable for heating applications.

**Question 10:**

How does the resistance of a wire vary with its area of cross-section?

**Answer:**

Resistance $R$ is inversely proportional to the cross-sectional area $A$ of the wire:

A wire with a larger cross-sectional area will have a lower resistance.

**Question 11:**

How will you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (a) 9 Ω, (b) 4 Ω?

**Answer:**

$\frac{1}{R_{\text{parallel}}} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \Rightarrow R_{\text{parallel}} = 3 \, \Omega$**(a) To get 9 Ω:**

Connect two 6 Ω resistors in parallel, their equivalent resistance is 3 Ω:Then connect this combination in series with the third 6 Ω resistor:

$R_{\text{total}} = 3 + 6 = 9 \, \Omega$

$R_{\text{series}} = 6 + 6 = 12 \, \Omega$**(b) To get 4 Ω:**

Connect two 6 Ω resistors in series, their equivalent resistance is 12 Ω:Then connect this combination in parallel with the third 6 Ω resistor:

$\frac{1}{R_{\text{total}}} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4} \Rightarrow R_{\text{total}} = 4 \, \Omega$

**Question 12:**

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

**Answer:**

Advantages of parallel connection:

- Each device gets the same voltage as the source.
- If one device fails, others continue to work.
- It allows for individual control of each device with a switch.
- Parallel circuits reduce the total resistance, allowing more current to flow.

**Question 13:**

Why does the cord of an electric heater not glow while the heating element does?

**Answer:**

The cord of an electric heater is made of materials with low resistance, which does not produce much heat when current flows through it. The heating element, however, has high resistance, causing it to heat up and glow.

**Question 14:**

What determines the rate at which energy is delivered by a current?

**Answer:**

The rate at which energy is delivered by a current is given by **electric power (P)**, which is calculated as:

where $V$ is voltage, $I$ is current, and $R$ is resistance.

**Question 15:**

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hours.

**Answer:**

$P = VI = 220 \times 5 = 1100 \, \text{W} = 1.1 \, \text{kW}$**Power of the motor:**

$\text{Energy} = \text{Power} \times \text{Time} = 1.1 \, \text{kW} \times 2 \, \text{hours} = 2.2 \, \text{kWh}$**Energy consumed in 2 hours:**So, the energy consumed is 2.2 kWh.