Why is long-distance power transmission done at high voltages?

Long-distance power transmission is done at high voltages because higher voltage results in lower current for the same power. This reduces heat losses in the transmission wires, as heat loss is proportional to the square of the current (H∝I2H∝I2)

What is electric energy, and how is it calculated?

Electric energy is the total energy supplied by a source of emf to maintain the electric current in a circuit over a given time. It is calculated using the formula: E = P × t = V × I × t Where: EE = Electric energy (in Joules) PP = Power (in Watts) tt = Time (in seconds) VV = Voltage (in Volts)

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Electric Energy and Joule's Law

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Electric Energy and Joule's Law

1.0Electric Energy

The total energy supplied by a source of emf in order to maintain the electric current in the circuit in a given time is called ‘electric energy’.

E = P × t = V × I × t

Unit of Electric Energy

SI unit : Joule

Here, 1 Joule = 1 volt-ampere-sec = 1 VAs

Other unit : 1 Watt-hour = 3600 J

1 Kilo Watt Hour (kWh) = 3.6 × 10⁶ J

Kilo Watt Hour (kWh) is a commercial unit of electric energy. If an electric device of 1 kW is used for 1hr, then the energy consumed is 1 kwh.

Long distance power transmission is done at high voltages because at high voltages, current flowing through the transmission wires is less. As a result, heat losses (H ∝ I²) will be less.

2.0Joule’s Law of Heating

The energy supplied to the circuit by a source (like a battery) in time t is given by,

E = P × t = V × I × t [ P = V × I]

This energy (E) gets dissipated in the resistor as heat. Thus, for a steady current I, the amount of heat H produced in time t is

H = V × I × t

Applying Ohm’s law, i.e., V= I × R, we get,

$H = I^{2} Rt$

This is known as Joule’s law of heating.

According to Joule’s law of heating, the heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor.

Example

1. Data for various electrical appliances is given in the table below.

Appliance	Power(W)	Potential difference (V)	Current (A)
Car headlamp	48	12	4
TV	240	240
Hairdryer		240	2
Iron	960	240
Kettle		240	10

The power for the hairdryer and current for the iron are, respectively :

(1) 480W ; 0.25 A

(2) 240W ; 4A

(3) 480W ; 4A

(4) 960 ; 2A

Solution

Option (3) is correct.

Given, for the hair dryer, Voltage (V) = 240 V, Current (I) = 2 amp.

Power of hair dryer

P = VI = 240 × 2 = 480 W.

Given, for the electric iron, Power (P) = 960 W, Voltage (V) = 240 V, current for the iron,

$I = \frac{P}{V} = \frac{960 W}{240 V} = 4 A$

3.0Also Read

Measurement of Time	Lenses	Motion & Its Type
Applications of Thermal Effects of Current	Optic Reflection of Light	Distance and Displacement
Optic Refraction of Light	Reflection From Spherical Mirrors	Velocity

Join ALLEN!

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I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Electric Energy and Joule's Law

1.0Electric Energy

The total energy supplied by a source of emf in order to maintain the electric current in the circuit in a given time is called ‘electric energy’.

E = P × t = V × I × t

Unit of Electric Energy

SI unit : Joule

Here, 1 Joule = 1 volt-ampere-sec = 1 VAs

Other unit : 1 Watt-hour = 3600 J

1 Kilo Watt Hour (kWh) = 3.6 × 10⁶ J

Kilo Watt Hour (kWh) is a commercial unit of electric energy. If an electric device of 1 kW is used for 1hr, then the energy consumed is 1 kwh.

Long distance power transmission is done at high voltages because at high voltages, current flowing through the transmission wires is less. As a result, heat losses (H ∝ I²) will be less.

2.0Joule’s Law of Heating

The energy supplied to the circuit by a source (like a battery) in time t is given by,

E = P × t = V × I × t [ P = V × I]

This energy (E) gets dissipated in the resistor as heat. Thus, for a steady current I, the amount of heat H produced in time t is

H = V × I × t

Applying Ohm’s law, i.e., V= I × R, we get,

$H = I^{2} Rt$

This is known as Joule’s law of heating.

According to Joule’s law of heating, the heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor.

Example

1. Data for various electrical appliances is given in the table below.

Appliance	Power(W)	Potential difference (V)	Current (A)
Car headlamp	48	12	4
TV	240	240
Hairdryer		240	2
Iron	960	240
Kettle		240	10

The power for the hairdryer and current for the iron are, respectively :

(1) 480W ; 0.25 A

(2) 240W ; 4A

(3) 480W ; 4A

(4) 960 ; 2A

Solution

Option (3) is correct.

Given, for the hair dryer, Voltage (V) = 240 V, Current (I) = 2 amp.

Power of hair dryer

P = VI = 240 × 2 = 480 W.

Given, for the electric iron, Power (P) = 960 W, Voltage (V) = 240 V, current for the iron,

$I = \frac{P}{V} = \frac{960 W}{240 V} = 4 A$

3.0Also Read

Measurement of Time	Lenses	Motion & Its Type
Applications of Thermal Effects of Current	Optic Reflection of Light	Distance and Displacement
Optic Refraction of Light	Reflection From Spherical Mirrors	Velocity

Frequently Asked Questions

Why is long-distance power transmission done at high voltages?

What is electric energy, and how is it calculated?

Frequently Asked Questions

Why is long-distance power transmission done at high voltages?

What is electric energy, and how is it calculated?

Join ALLEN!

Electric Energy and Joule's Law

1.0Electric Energy

Unit of Electric Energy

2.0Joule’s Law of Heating

3.0Also Read

Table of Contents

Join ALLEN!

Electric Energy and Joule's Law

1.0Electric Energy

Unit of Electric Energy

2.0Joule’s Law of Heating

3.0Also Read

Table of Contents