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Electrical Energy and Power



Heating Effect Of Electric Current ($ H = I^2Rt $)

When an electric current flows through a conductor, especially one offering resistance, electrical energy is converted into heat energy. This phenomenon is known as the heating effect of electric current or Joule heating. It occurs because the charge carriers (electrons in metals) collide with the atoms or ions of the conductor's lattice as they drift through the material. These collisions transfer kinetic energy from the drifting electrons to the lattice atoms, increasing their vibrational energy, which is perceived as heat.


Joule's Law of Heating

James Prescott Joule experimentally found that the heat produced in a resistor is:

  1. Directly proportional to the square of the current ($I^2$) flowing through the resistor.
  2. Directly proportional to the resistance ($R$) of the resistor.
  3. Directly proportional to the time ($t$) for which the current flows.

Combining these observations, Joule formulated his law of heating:

$ \text{Heat produced } (H) \propto I^2 R t $

Introducing a constant of proportionality (which is 1 in SI units where Heat is measured in Joules), we get the formula for the heat produced:

$ H = I^2 R t $


Derivation of the Heating Formula

Let's derive this formula from basic principles. Consider a conductor with resistance $R$ through which a current $I$ is flowing for a time $t$. Let the potential difference across the conductor be $V$.

By definition, potential difference is the work done per unit charge ($V = W/Q$). Thus, the work done ($W$) or energy supplied by the source in moving a charge $Q$ across a potential difference $V$ is:

$ W = V Q $

We also know that electric current is the rate of flow of charge ($I = Q/t$). So, the charge $Q$ that flows in time $t$ is $Q = I t$.

Substituting $Q = It$ into the expression for work done:

$ W = V (I t) $

$ W = V I t $

This work done by the source is converted into heat in the resistor (assuming no other energy conversion happens, like mechanical work by a motor). So, the heat produced $H$ is equal to the work done $W$:

$ H = V I t $

Now, we can use Ohm's Law ($V=IR$) to express the heat produced in terms of current and resistance, or voltage and resistance.

Substituting $V = IR$ into $H = VIt$:

$ H = (IR) I t $

$ H = I^2 R t $

This is the formula for heat produced mentioned in the subheading.

Alternatively, substituting $I = V/R$ (from Ohm's Law) into $H = VIt$:

$ H = V \left(\frac{V}{R}\right) t $

$ H = \frac{V^2}{R} t $

All three formulas ($H = VIt$, $H = I^2Rt$, and $H = V^2t/R$) are equivalent and represent the electrical energy converted into heat. The SI unit for heat (and energy) is the joule (J).


Example 1. A heating element of an electric iron has a resistance of 50 $\Omega$. If it draws a current of 4 A when connected to a 220 V supply, calculate the heat generated in 10 minutes.

Answer:

Given:

Resistance, $R = 50 \, \Omega$

Current, $I = 4 \, A$

Time, $t = 10 \, \text{minutes} = 10 \times 60 \, \text{seconds} = 600 \, s$

Supply Voltage, $V = 220 \, V$ (Note: We can check if the given voltage and current are consistent with Ohm's law for this resistance: $V = IR = 4\,A \times 50\,\Omega = 200\,V$. There seems to be a slight inconsistency in the example data, as 4A through 50 Ohms implies 200V, not 220V. Let's use the given current and resistance as these directly allow calculation using $I^2Rt$).

We use the formula for heat generated: $H = I^2 R t$

Substitute the values:

$ H = (4 \, A)^2 \times (50 \, \Omega) \times (600 \, s) $

$ H = 16 \, A^2 \times 50 \, \Omega \times 600 \, s $

$ H = 16 \times 50 \times 600 \, J $

$ H = 800 \times 600 \, J $

$ H = 480000 \, J $

$ H = 4.8 \times 10^5 \, J $

The heat generated in the heating element in 10 minutes is 480,000 Joules or 480 kJ.


Practical Applications Of Heating Effect Of Electric Current

The heating effect of electric current is utilised in numerous daily applications where electrical energy is converted into heat for practical purposes.

  1. Electric Heating Appliances: Devices like electric irons, room heaters, water heaters (geysers), toasters, electric kettles, and electric ovens use heating elements made of high-resistance materials like Nichrome alloy. When current passes through these elements, they become very hot and radiate heat.

  2. Electric Bulb (Incandescent Lamp): In an incandescent bulb, a thin filament (usually made of tungsten, which has a very high melting point) is heated to a very high temperature (around 2500-3000°C) by the passage of current. At this high temperature, the filament glows and emits light. However, a significant portion of the electrical energy is wasted as heat, making them less energy-efficient compared to modern lighting like LEDs.

  3. Electric Fuse: A fuse is a safety device in electrical circuits. It consists of a wire made of a material with a low melting point (e.g., an alloy of lead and tin) placed in a ceramic cartridge. It is connected in series with the live wire of the circuit. If the current in the circuit exceeds a safe limit (due to overload or short circuit), the fuse wire gets heated up significantly ($H \propto I^2$) and melts, breaking the circuit and preventing damage to appliances or potential fire hazards. Fuses are rated by the maximum current they can safely carry.

    Diagram showing an electric fuse in a circuit

    An electric fuse acting as a safety device in a circuit.

  4. Electric Welding: In resistance welding, a large current is passed through the metals to be joined. The resistance at the contact point generates enough heat to melt the metal and fuse them together.



Electric Power ($ P = VI = I^2R = V^2/R $)

In electricity, electric power is the rate at which electrical energy is transferred or converted from one form to another in an electric circuit. It represents how quickly work is being done or energy is being consumed or supplied.

Just like mechanical power is the rate of doing mechanical work, electric power is the rate of doing electrical work or the rate at which electrical energy is dissipated (converted to heat or other forms) or supplied.


Definition and Basic Formula

Power ($P$) is defined as the rate at which energy ($W$) is transferred or consumed.

$ P = \frac{W}{t} $

From the derivation of the heating effect, we found that the work done or energy transferred in an electric circuit is $W = VIt$.

Substituting this expression for $W$ into the power definition:

$ P = \frac{VIt}{t} $

$ P = V I $

This is the basic formula for electric power. It states that the power is equal to the product of the potential difference across the component and the current flowing through it.


Units of Electric Power

The SI unit of power is the watt (W). It is named after James Watt, the inventor of the steam engine.

From the formula $P = VI$, the unit of watt can be defined in terms of volt and ampere:

$ 1 \text{ Watt} = 1 \text{ Volt} \times 1 \text{ Ampere} $

One watt is the power consumed by a device that carries a current of 1 ampere when a potential difference of 1 volt is applied across it.

Larger units of power include kilowatt ($kW = 10^3 W$), megawatt ($MW = 10^6 W$), etc.


Other Formulas for Electric Power

Using Ohm's Law ($V=IR$), we can express power in different forms, which are useful depending on the known quantities (Voltage, Current, or Resistance).

  1. Power in terms of Current and Resistance ($P = I^2R$)

    Start with the basic formula $P = VI$. Substitute $V = IR$ (from Ohm's Law):

    $ P = (IR) I $

    $ P = I^2 R $

    This formula is particularly useful when you know the current flowing through a resistor and its resistance, and you want to find the power dissipated as heat (Joule heating).

  2. Power in terms of Voltage and Resistance ($P = V^2/R$)

    Start with the basic formula $P = VI$. Substitute $I = V/R$ (from Ohm's Law):

    $ P = V \left(\frac{V}{R}\right) $

    $ P = \frac{V^2}{R} $

    This formula is useful when you know the potential difference across a component and its resistance, and you want to find the power. This is often used for appliances designed to operate at a specific voltage (like 220 V in India). The resistance of the appliance is usually fixed, so the power rating is determined by the operating voltage.

So, the three common formulas for electric power are:

$ P = V I = I^2 R = \frac{V^2}{R} $

These are equivalent expressions derived using Ohm's Law.


Example 2. An electric bulb is rated 100 W, 220 V. (a) What is the resistance of the bulb? (b) What is the current drawn by the bulb when connected to a 220 V supply?

Answer:

Given rating of the bulb: Power $P = 100 \, W$, Voltage $V = 220 \, V$. The rating indicates the power consumed when operated at the specified voltage.

(a) To find the resistance ($R$) of the bulb, we can use the formula $P = V^2/R$, as we know $P$ and $V$.

Rearrange the formula to find $R$: $R = \frac{V^2}{P}$

Substitute the given values:

$ R = \frac{(220 \, V)^2}{100 \, W} = \frac{220 \times 220}{100} \, \Omega $

$ R = \frac{48400}{100} \, \Omega = 484 \, \Omega $

The resistance of the bulb is 484 $\Omega$. (Assuming the resistance is constant, which is an approximation for a filament bulb, but standard for such problems).

(b) To find the current ($I$) drawn by the bulb when connected to a 220 V supply, we can use the formula $P = VI$.

Rearrange the formula to find $I$: $I = \frac{P}{V}$

Substitute the given values:

$ I = \frac{100 \, W}{220 \, V} = \frac{10}{22} \, A = \frac{5}{11} \, A $

$ I \approx 0.455 \, A $

The current drawn by the bulb is approximately 0.455 A.

Alternatively, we could use Ohm's Law $I = V/R$ with the calculated resistance from part (a):

$ I = \frac{220 \, V}{484 \, \Omega} = \frac{220}{484} \, A = \frac{110}{242} \, A = \frac{55}{121} \, A = \frac{5}{11} \, A \approx 0.455 \, A $

Both methods give the same result.



Electrical Energy, Power

We have discussed electric power as the rate at which electrical energy is consumed or supplied. Now, let's explicitly define electrical energy and its common units, particularly the unit used for commercial purposes.


Electrical Energy (W or E)

Electrical energy is the total work done by the electric current in a circuit over a period of time. It is the amount of energy converted from electrical form to other forms (like heat, light, mechanical energy, etc.) by the components in the circuit.

Since power ($P$) is the rate of energy transfer ($W/t$), the total electrical energy ($W$) transferred or consumed over a time interval $t$ can be calculated by multiplying the power by the time:

$ \text{Energy } (W) = \text{Power } (P) \times \text{Time } (t) $

$ W = P t $

Using the different formulas for power ($P = VI = I^2R = V^2/R$), we can also write the formulas for electrical energy consumed:

$ W = V I t $

$ W = I^2 R t $ (This is the same as the heat formula $H = I^2Rt$, because in a pure resistor, all electrical energy is converted to heat)

$ W = \frac{V^2}{R} t $


Units of Electrical Energy

The SI unit of electrical energy is the joule (J). One joule is the energy consumed or transferred when 1 watt of power is used for 1 second.

$ 1 \text{ Joule} = 1 \text{ Watt} \times 1 \text{ Second} $


Commercial Unit of Electrical Energy (Kilowatt-hour)

The joule is a very small unit for practical purposes, especially when dealing with the amount of electricity consumed in households or industries over extended periods (like a month). Therefore, a larger, more convenient unit is used for commercial measurement of electrical energy. This unit is the kilowatt-hour (kWh).

One kilowatt-hour is the energy consumed when a device having a power of 1 kilowatt (kW) operates for 1 hour (h).

$ 1 \text{ kWh} = 1 \text{ kilowatt} \times 1 \text{ hour} $

Let's convert kilowatt-hour into joules:

$ 1 \text{ kW} = 1000 \text{ W} $

$ 1 \text{ hour} = 60 \text{ minutes} = 60 \times 60 \text{ seconds} = 3600 \text{ s} $

$ 1 \text{ kWh} = (1000 \text{ W}) \times (3600 \text{ s}) $

$ 1 \text{ kWh} = 3600000 \text{ Ws} = 3.6 \times 10^6 \text{ J} $

So, 1 kilowatt-hour is equal to $3.6 \times 10^6$ joules or 3.6 megajoules (MJ).

In common language, 1 kilowatt-hour is often referred to as "1 unit" of electricity consumption, as seen in electricity bills.


Example 3. An electric refrigerator rated 400 W operates 8 hours per day. Calculate the cost of the energy to operate it for 30 days at ₹3.00 per kWh.

Answer:

Given:

Power rating of refrigerator, $P = 400 \, W$

Operating time per day $= 8 \, hours$

Number of days $= 30$ days

Cost per kWh $= ₹3.00$

First, calculate the total time the refrigerator operates in 30 days:

Total time $= 8 \, \text{hours/day} \times 30 \, \text{days} = 240 \, \text{hours} $

Now, calculate the total electrical energy consumed in kWh. Power is in Watts, so convert it to kilowatts:

$ P = 400 \, W = \frac{400}{1000} \, kW = 0.4 \, kW $

Total energy consumed ($W$) in kWh is Power (kW) $\times$ Total time (h):

$ W = P \times t = (0.4 \, kW) \times (240 \, h) $

$ W = 96 \, kWh $

The total energy consumed is 96 kilowatt-hours (or 96 units).

Now, calculate the total cost:

Total Cost $= \text{Total Energy Consumed (kWh)} \times \text{Cost per kWh} $

Total Cost $= 96 \, kWh \times ₹3.00/\text{kWh} $

Total Cost $= ₹288.00 $

The cost to operate the refrigerator for 30 days is ₹288.