What Is Electrical Energy?

Electrical energy is the energy derived from the electric potential energy or kinetic energy of the charged particles. In general, it is referred to as the energy that has been converted from electric potential energy. We can define electrical energy as the energy generated by the movement of electrons from one point to another.

A cell has two terminals – a negative and a positive terminal. The negative terminal has the excess of electrons whereas the positive terminal has a deficiency of electrons. Let us take the positive terminal as \(A\) and the electrical potential at \(A\) is given by \(V ( A )\). Similarly, the negative terminal is \(B\) and the electrical potential at \(B\) is given by \(V ( B )\). Electric current flows from \(A\) to \(B\) , and thus \(V ( A )> V ( B )\).
The potential difference between \(A\) and \(B\) is given by
\(
V = V ( A )- V ( B )>0
\)
Mathematically, electric current is defined as the rate of flow of charge through the cross-section of a conductor. If the two charges have opposite signs, then the potential energy is negative. This means that the potential energy decreases when the two charges move closer together (the potential energy will get more negative). A negative potential energy means that work must be done against the electric field in moving the charges apart.

In a time interval \(\Delta t\), an amount of charge \(\Delta Q=I \Delta t\) travels from \(A\) to \(B\). The potential energy of the charge at \(A\), by definition, was \(Q V(A)\) and similarly at \(B\), it is \(Q V(B)\). Thus, change in its potential energy \(\Delta U_{\text {pot }}\) is
\(
\begin{aligned}
& =\Delta Q[(V(B)-V(A)]=-\Delta Q V  \\
& =-I V \Delta t<0
\end{aligned}
\)
If we take the kinetic energy of the system into account, it would also change if the charges inside the conductor moved without collision. This is to keep the total energy of the system unchanged. Thus, by conservation of total energy, we have:
\(
\Delta K=-\Delta \text { Upot }
\)
Or \(\Delta K=I V \Delta t>0\)
Thus, in case charges were moving freely through the conductor under the action of the electric field, their kinetic energy would increase as they move.
Thus, in case charges were moving freely through the conductor under the action of the electric field, their kinetic energy would increase as they move.

We have, however, seen earlier that on average, charge carriers do not move with acceleration but with a steady drift velocity. This is because of the collisions with ions and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms. The atoms vibrate more vigorously, i.e., the conductor heats up. Thus, in an actual conductor, an amount of energy dissipated as heat in the conductor during the time interval \(\Delta t\) is,
\(
\Delta W=I V \Delta t
\)
The energy dissipated per unit time is the power dissipated \(P=\Delta W / \Delta t\) and we have,
\(P=I V\)

Electrical Power

What is Electrical Power?

Every electrical equipment or device that we use has specific power ratings mentioned on it. It means it consumes that specific rated power & converts that electrical power to good use. Such as a cellphone, utilizes the power from the battery, to power its display unit (converting it into light), speakers (for audio generation) & its processors (for logical operations), etc. Same as machines that consume electrical power & generate mechanical power & heaters generate heat energy.

Definition: Generally, the definition of power is the rate of energy transferred or the energy transferred in a unit time. So according to the definition, the electrical power is the rate of flow of electrical energy or the work done on electrical charges in an electrical circuit.

Energy liberated per second in a device is called its power. The electrical power \(P\) delivered or consumed by an electrical device is given by \(P = VI\), where \(V =\) Potential difference across the device and \(I =\text { Current. }\)

If the current enters the higher potential point of the device then electric power is consumed by it (i.e. acts as load). If the current enters the lower potential point then the device supplies power (i.e. acts as a source).

Mathematically, if \(W\) joules is the amount of work done in an electric circuit in a time of \(t\) seconds. Then, the electric power in the circuit is given by,
Power is the rate at which that work is done.
\(
P=\frac{W}{t}
\)
In differential form, instantaneous power
\(
p(t)=\frac{ d w(t)}{ d t}=\vec F \cdot \frac{d \vec s }{d t}= \vec F \cdot \vec v
\)
For electrical energy to move electrons and produce a flow of current around a circuit, work must be done, that is the electrons must move by some distance through a wire or conductor. The work done is stored in the flow of electrons as energy. Thus “Work” is the name we give to the process of energy.
Energy \(=\int p d t\)

Power consumed by a resistor (resistor is a passive circuit element that dissipates energy)
\(
P = VI =I ^2 R =\frac{ V ^2}{ R } \text { using Ohm’s law }
\)

When a current is passed through a resistor energy is wasted in overcoming the resistance of the wire. This energy is converted into heat.
\(
W = VIt = I ^2 Rt =\frac{ V ^2}{ R } t
\)

Electrical Heating

Cause of Heating

The potential difference applied across the two ends of a conductor sets up an electric field. Under the effect of the electric field, electrons accelerate and as they move, they collide against the ions and atoms in the conductor, the energy of electrons transferred to the atoms and ions appears as heat.

Joules’s Law of Heating

When a current \(I\) is made to flow through an ohmic resistance \(R\) for time \(t\), heat \(H\) is produced such that
\(
H = I ^2 R t= P \times t = VIt =\frac{ V ^2}{ R } t \text { Joule }=\frac{I^2 Rt }{4.2} \text { Calorie }
\)
The heat produced in a conductor does not depend upon the direction of the current.
\(
\begin{array}{ll}
\text { SI unit : joule } ; & \text { Practical Units : } 1 \text { kilowatt hour }( kWh ) \\
1 kWh =3.6 \times 10^6 \text { joules }=1 \text { unit }; & 1 \text { BTU (British Thermal Unit })=1055 J
\end{array}
\)

Example 24: If the bulb rating is 100 watt and 220 V then determine
(a) Resistance of filament
(b) Current through filament
(c) If the bulb operates at 110 volt power supply then find the power consumed by the bulb.

Solution: The bulb rating is 100 W and 220 V bulb means when 220 V potential difference is applied between the two ends then the power consumed is 100 W
Here
\(
\begin{aligned}
& V =220 Volt \\
& P =100 W \\
& \frac{V^2}{R}=100
\end{aligned}
\)
So, \(R=484 \Omega\)
Since Resistance depends only on material hence it is constant for bulb
\(
I =\frac{V}{R}=\frac{220}{22 \times 22}=\frac{5}{11} Amp .
\)
power consumed at 110 V
\(
\therefore \text { power consumed }=\frac{110 \times 110}{484}=25 W
\)

Example 25: Estimate the cost of cooking a turkey for 4 h in an oven that operates continuously at 20.0 A and 240 V (estimated price of \(8.00\) cent per kilowatt hour).

Solution: The power used by the oven is
\(
P =I V=(20.0 A)(240 V)=4800 W=4.80 kW
\)
Because the energy consumed equals power \(\times\) time, the amount of energy for which you must pay is
\(
\text { Energy }= P t=(4.80 kW)(4 h)=19.2 kWh
\)
If the energy is purchased at an estimated price of \(8.00\) cents per kilowatt hour, the cost is
\(
\text { Cost }=(19.2 kWh )(0.080 / kWh )=$1.54
\)

Example 26: A light bulb is connected to a battery in a series circuit. Explain the change in brightness of the light bulb if an identical light bulb is added to the circuit in series.

Solution: Adding an identical light bulb in series doubles the total resistance of the circuit. This halves the current flowing through the light bulbs \((V=I R)\). It also causes each light bulb to receive half as much energy from the battery; the voltage across each light bulb is halved.
Since electrical power is given by:
\(
\begin{gathered}
P=V I \\
P_{\text {new }}=\left(\frac{V}{2}\right)\left(\frac{I}{2}\right)=\frac{P}{4}
\end{gathered}
\)
The power dissipated in each light bulb is reduced by a factor of 4. So brightness gets reduced (as you add more series resistor in a series circuit the resistance increases, current decreases, and voltage also decreases across each resistor, so brightness gets reduced).

Power in Series and Parallel Circuits

Power is a measure of the rate of work. Per the physics law of conservation of energy, the power dissipated in the circuit must equal the total power applied by the source(s). Therefore, an interesting rule for total circuit power versus individual component power is that it is additive for any circuit configuration as shown in the table below.

Power in Series Circuits

Whether a resistor is connected in series or in parallel, the power dissipated in the resistor is:
\(
\begin{aligned}
& P_1=V_1 I \\
& P_1=\frac{V_1^2}{R_1} \\
& o r \\
& P_1=I^2 R_1 \\
& P_2=I^2 R_2 \\
& P_3=I^2 R_3 \\
\end{aligned}
\)
The total power output from the battery is, of course,
\(
\begin{aligned}
& P=V I=I\left(V_1+V_2+V_3\right)=I V_1+I V_2 +I V_3 \\
&  \text { Total Power from battery } P=P_1+P_2+P_3 \\
\end{aligned}
\)
For any parallel (or series) combination of \(n\) resistors, the Equation would be:
\(
P=P_1+P_2+P_3+\ldots+P_n
\)

• \(P_{\text {consumed }}\) (brightness) \(\propto V \propto R \propto \frac{1}{P_{\text {rated }}}\), i.e. in series combination bulb of lesser wattage will give more bright light and potential difference appearing across it will be more.

Power in Parallel Circuits

Whether a resistor is connected in series or in parallel, the power dissipated in the resistor is:
\(
\begin{aligned}
& P_1=V I_1 \\
& P_1=\frac{V^2}{R_1} \\
& o r \\
& P_1=I_1^2 R_1
\end{aligned}
\)
The power dissipated in \(R _2\) is calculated in a similar way (\(P_2=V I_2\)) and across \(R _3\) is (\(P_3=V I_3\)). The total power output from the battery is, of course,
\(
\begin{aligned}
& P=V I=V\left(I_1+I_2+I_3\right)=V I_1+V I_2 +V I_3 \\
&  \text { Total Power from battery } P=P_1+P_2+P_3 \\
\end{aligned}
\)
For any parallel (or series) combination of \(n\) resistors, the Equation would be:
\(
P=P_1+P_2+P_3+\ldots+P_n
\)

\(P_{\text {consumed }}\left(\right.\) brightness) \(\propto P_{\text {rated }} \propto I \propto \frac{1}{R}\), i.e. in parallel combination, bulb of greater wattage will give more bright light and more current will pass through it.

Table method for series circuits—power is additive.

\(
\begin{array}{|c|c|c|c|c|c|}
\hline & R _1 & R _2 & R_3 & \text { Total } & \text { Units } \\
\hline \text { V } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Add } & \text { V } \\
\hline \text { I } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Equal } & \text { A } \\
\hline \text { R } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Add } & \Omega \\
\hline \text { P } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Add } & \text { w } \\
\hline
\end{array}
\)

Table method for Parallel circuits—power is additive.

\(
\begin{array}{|c|c|c|c|c|c|}
\hline & R _1 & R _2 & R_3 & \text { Total } & \text { Units } \\
\hline \text { V } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Equal } & \text { V } \\
\hline \text { I } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Add } & \text { A } \\
\hline \text { R } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Reduce } & \Omega \\
\hline \text { P } & \longrightarrow & \longrightarrow & \longrightarrow & \text { Add } & \text { w } \\
\hline
\end{array}
\)

Note: Power is additive in any configuration of a resistive circuit-series, parallel, or series-parallel
\(P_{\text {total }}=P_1+P_2+\ldots P_n\)