Electricity-Notes
Physics - Notes
Electricity
ELECTRIC CURRENT AND CIRCUIT
Electric Circuit
An electric circuit is a closed, continuous path through which electric current can flow. A simple circuit usually includes:- a source of electrical energy (cell, battery or power supply)
- conducting wires
- a switch to open or close the path
- a device or load such as a bulb or resistor
Direction of Current
The conventional direction of current is taken from the positive terminal of the cell or battery to the negative terminal, even though electrons actually move in the opposite direction.
Representation of Electric Circuits
For convenience, electric circuits are often shown using circuit diagrams, where standard symbols are used for components like:
- Cell
- Battery
- Switch
- Bulb
- Resistor
- Ammeter
- Voltmeter
Circuit diagrams make it easier to study and understand the behaviour of electricity in different arrangements.
Electric Current
If a net charge \(Q\), flows across any cross-section of a conductor in time \(t\), then the current \(I\), through the cross-section is
\[\boxed{I=\frac{Q}{t}}\]The SI unit of electric charge is coulomb (C), which is equivalent to the charge contained in nearly \(6 × 10^{18}\) electrons.
The electric current is expressed by a unit called ampere (A), named after the French scientist, Andre-Marie Ampere (1775–1836).
One ampere is constituted by the flow of one coulomb of charge per second
\[\boxed{\mathrm{1\,A=\frac{1\,C}{1\,s}}}\]
Small quantities of current are expressed in milliampere
\[\mathrm{1\,mA = 10^{–3} A}\]
or in microampere \[\mathrm{1\, \mu A = 10^{–6} A}\]
Example-1
A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
Solution
\(\mathrm{I}\)=0.5A
time \(\mathrm{t}\)=10 minutes = \(\mathrm{10\times 60=600\,s}\)
Example-2
An electric bulb draws a current of 0.25 A for 20 minutes. Calculate the amount of electric charge that flows throught he circuit.
Solution
\(\mathrm{I}\)=0.25Atime \(\mathrm{t}\)=10 minutes = \(\mathrm{20\times 60=1200\,s}\) \[ \begin{aligned} I&=\frac{Q}{t}\\\\ \implies Q&=I\times t\\\\ &=0.25\times 1200\\ &=300\,C \end{aligned} \] Thus, the amount of electric charge that flows through the circuit is 300 C.
Summary
- Electric current = rate of flow of charge.
- Current flows only when the circuit is closed.
- A circuit contains source, connecting wires, switch and load.
- Direction of conventional current is from + to – terminal.
- Circuit diagrams use standard symbols for components.
ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE
Electric potential
Electric potential is the amount of work needed to bring a unit positive charge from infinity (or a
reference point) to a specific point in an electric field.
It tells us how strongly a point in the circuit
can pull or push electric charges. In simple words, electric potential is the “electric pressure” that
allows charges to move.
Electric potential is represented by V and its SI unit is the volt (V).
Electric Potential Difference (Voltage)
The electric potential difference between two points is defined as the amount of work done to move a unit
positive charge from one point to the other.
It is the actual driving force that makes electric charges flow through a conductor.
\[
\boxed{\mathrm{V=\frac{W}{Q}}}
\]
Where,
\(\mathrm{V}\) = potential difference
\(\mathrm{W}\) = work done
\(\mathrm{Q}\)= charge
If one joule of work is required to move one coulomb of charge, the potential difference is one
volt.
The SI unit of electric potential difference is volt (V), named after
Alessandro Volta (1745–1827), an Italian physicist.
One volt is the potential difference between two points in a current carrying conductor
when 1 joule of work is done to move a charge of 1 coulomb from one
point to the other.
Therefore,
\begin{align*}
1\, V &= \frac{1\, \text{Joule}}{1\, \text{Coulomb}} \\\\
1\, V &= 1\, J\,C^{-1}
\end{align*}
The potential difference is measured by means of an instrument called
the voltmeter.
The voltmeter is always connected in parallel across the
points between which the potential difference is to be measured.
Summary
- Electric potential tells how much work is needed to bring a unit charge to a point.
- Potential difference is the work done in moving a unit charge between two points.
- It is measured in volts.
- A cell or battery provides potential difference.
- A voltmeter is used to measure it and must be connected in parallel.
Example-3
How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
Solution
\(\mathrm{V}=\) 12 V
\(\mathrm{Q}=\) 2 C
Work Done \(\mathrm{W}=\)
Example-4
How much work is done in moving a charge of 2 C from a point at 118 Volts to a point of 128 Volts?
solution
\(\mathrm{V}=\) 128 V -118 V = 10 V
\(\mathrm{Q}=\) 2 C
Work Done \(\mathrm{W}=\)
\[
\begin{aligned}
V&=\frac{W}{Q}\\\\
\implies W&=V\times Q\\
&=10\times 2\\
&=20\,\mathrm{J}
\end{aligned}
\]
Ohm’s Law
Mathematically, \[\begin{aligned}V& \propto I\\ V&=IR\end{aligned}\]
Where,
V = potential difference (in volts),
I = current (in amperes),
R = resistance of the conductor (in ohms).
The constant of proportionality \(R\) is called resistance, which measures how strongly the conductor
opposes current flow.
In 1827, a German physicist Georg Simon Ohm (1787–1854) found out the relationship between the current I, flowing in a metallic wire and the potential difference across its terminals.
If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is \(1\,\Omega\) \[ 1\Omega=\frac{1\,\text{Volt}}{1\, \text{ampere}} \] \[\boxed{I=\frac{V}{R}}\]
FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS
Resistance is the property of a conductor that opposes the flow of electric current. The resistance of a conductor varies with several physical factors. These factors are explained below:
1. Length of the Conductor (L)
Resistance increases with the increase in the length of the wire. A longer wire causes more collisions of electrons with atoms, increasing opposition to current flow.
\[ R \propto l\tag{1} \]
2. Area of Cross-Section (A)
Resistance decreases when the thickness or area of cross-section of the conductor increases. A thicker wire provides a wider path for electrons.
\[ R \propto \frac{1}{A}\tag{2} \]
Combining Eqs. (1) and (2) we get \[R\propto \frac{l}{A}\] or, \[\boxed{R=\rho\frac{l}{A}}\]
where \(\rho\) (rho) is a constant of proportionality and is called the electrical
resistivity of the material of the conductor.
The SI unit of resistivity is
\(\Omega\,m\).
Electrical resistivity* of some substances at 20°C
| Substance | Material | Resistivity (\(\Omega\,m\)) |
|---|---|---|
| Conductors | Silver | \(1.60 × 10^{–8}\) |
| Copper | \( 1.62 × 10^{–8}\) | |
| Aluminium | \( 2.63 × 10^{–8}\) | |
| Tungsten | \(5.20 × 10^{–8}\) | |
| Nickel | \(6.84 × 10^{–8}\) | |
| \(Iron\) | \(10.0 × 10^{–8}\) | |
| Chromium | \( 12.9 × 10^{–8}\) | |
| Mercury | \( 94.0 × 10^{–8}\) | |
| Manganese | \(1.84 × 10^{–6}\) | |
| Alloys | Constantan (alloy of Cu and Ni) | \(49 × 10^{–6}\) |
| Manganin (alloy of Cu, Mn and Ni) | \(44 × 10^{–6}\) | |
| Nichrome (alloy of Ni, Cr, Mn and Fe) | \(100 × 10^{–6}\) | |
| Insulators | Glass | \(10^{10} – 10^{14}\) |
| Hard rubber | \( 10^{13} – 10^{16}\) | |
| Ebonite | \(10^{15} – 10^{17}\) | |
| Diamond | \(10^{12} - 10^{13}\) | |
| Paper (dry) | \(10^{12}\) |
3. Nature (Material) of the Conductor
Different materials offer different resistances. Metals such as copper and aluminium have low resistance, while alloys like nichrome and manganin have high resistance.
4. Temperature of the Conductor
For metallic conductors, resistance increases with temperature because atoms vibrate faster, causing more collisions with electrons.
Summary
- Resistance ∝ length (longer wire → more resistance)
- Resistance ∝ 1/area (thicker wire → less resistance)
- Resistance depends on the material of the conductor
- Resistance increases with temperature in metals
Example-5
(a) How much current will an electric bulb draw from a 220 V source,
if the resistance of the bulb filament is 1200 Ω?
(b) How much
current will an electric heater coil draw from a 220 V source, if
the resistance of the heater coil is 100 Ω?
Solution
(a) Voltage Source =220 V
Resistance of filament (R)=1200 \(\Omega\)
Current I
(b) Voltage Source =220 V
Resistance of filament (R)=100 \(\Omega\)
Current I
Eample-6
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Solution
Potential Difference between terminals=60 V
Current Drawn=4 A
Let Resistance of Heater R
New Potential difference V=120
\(R=15\,\Omega\)
Example-7
Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.
Solution
Resistance of Wire=26 \(\Omega\)
Length of the Wire \(l\)=1 m
Diameter of the Wire =0.3 m
\(\implies\) radius of wire = \(\frac{0.3}{2}=0.15\times 10^{-3}\,m\)
The resistivity of the metal at 20°C is \(1.85 × 10^{–6} Ω\, m\).
From
Table given above, we see that this is the resistivity of manganese.
Example-8
Solution
Length of the wire = l
Area of Cross section = A
Resistance of wire = \(4\Omega\)
Let Resistance of new wire be \(R_2\) and
length of the wire = l/2
Cross section of the wire = 2A
Substituting value of \(\rho\) from Eqn (1)
$$\begin{aligned} R_2&=\dfrac{4\cdot A}{l}\cdot \dfrac{l}{4A}\\\\ &=1\Omega \end{aligned}$$The resistance of the new wire is \(1\,\Omega\).
RESISTANCE OF A SYSTEM OF RESISTORS
When more than one resistor is connected in an electric circuit, the overall or equivalent resistance of the combination depends on the way each resistor is arranged. Resistors can be connected mainly in two ways:
1. Resistors in Series
Resistors are connected in series when they are joined end-to-end and the same current flows through all of them.
\[ R_{\text{series}} = R_1 + R_2 + R_3 + \cdots \]
Derivation
Potential difference V in the given circuit is equal to the sum of potential differences \(V_1, V2_, and V_3\). That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is,
\[ V=V_1 + V_2 + V_3\tag{1} \] In the electric circuit shown in given figure , let \(I\) be the current through the circuit. The current through each resistor is also \(I\). It is possible to replace the three resistors joined in series by an equivalent single resistor of resistance \(R_s\), such that the potential difference \(V\) across it, and the current \(I\) through the circuit remains the same. Applying the Ohm’s law to the entire circuit, we have \[V=IR\tag{2}\] On applying Ohm’s law to the three resistors separately, we further have \[ \begin{align} V_1=IR_1\\ V_2=IR_2\\ V_3=IR_3 \end{align}\] From Eqn (1) and (2) \[\begin{aligned} IR&=IR_1 + IR_2 + IR_3 \end{aligned}\] \[\boxed{R_s=R_1+R_2+R_3}\]In a series circuit, electrons pass through each resistor one after another. The total opposition increases, making the equivalent resistance greater than any individual resistor.
- Current remains the same through each resistor.
- Voltage divides across the resistors.
2. Resistors in Parallel
Resistors are in parallel when each resistor is connected across the same two points, giving multiple paths for current.
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \]
Make a parallel combination, XY, of three resistors having resistances \(R_1,\, R_2,\text{ and
}R_3\), respectively. Connect it with a battery, a plug key and an ammeter, as shown in
given figure (Resistors in Parallel). Also connect a voltmeter in parallel with the combination of
resistors.
Let the current be \(I\). Also take the voltmeter reading. It gives the potential difference \(V\),
across the combination.
Total current \(I\), is equal to the sum of the
separate currents through each branch of the combination.
\[
I=I_1+I_2+I_3\tag{1}
\]
Let \(R_p\) be the equivalent resistance of the parallel combination of
resistors.
By applying Ohm’s law to the parallel combination of resistors, we have
\[I=\frac{V}{R_p}\tag{2}\]
On applying Ohm’s law to each resistor, we have
\[
\begin{aligned}
I_1&=\frac{V}{R_1}\\
I_2&=\frac{V}{R_2}\\
I_3&=\frac{V}{R_3}
\end{aligned}
\]
From Eqs. (1) to (2), we have
\[
\frac{V}{R_p}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}
\]
or,
\[
\boxed{\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}
\]
In a parallel arrangement, current divides among the branches. Because several independent paths are available, the total resistance becomes less than the smallest individual resistor.
- Voltage remains the same across each resistor.
- Current splits in different branches.
3. Importance of Combining Resistors
Understanding series and parallel combinations is essential for designing circuits, controlling current, managing heating effects, and ensuring safe operation of electrical systems.
Summary
- Series: \( R_{\text{eq}} = R_1 + R_2 + R_3 \); current same; voltage divides; resistance increases.
- Parallel: \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \); voltage same; current divides; resistance decreases.
Example-9
An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω
resistance are connected to a 6 V battery (Fig. 11.9).
Calculate
(a) the
total resistance of the circuit,
(b) the current through the circuit, and
(c)
the potential difference across the electric lamp and conductor.
Solution
Resistance of Lamp \(R_l=20\,\Omega\)
Conductor Resistance \(R_c=4\,\Omega\)
Voltage \(V=6\,V\)
(a) Total Resistance \(R\)
\[
\begin{aligned}
R&=R_l+R_c\\
R&=20+4\\
&=24\,\Omega
\end{aligned}
\]
(b) Cutrrent through the circuit \(I\)
\[
\begin{aligned}
I&=\frac{V}{R}\\
&=\frac{6}{24}\\
&=0.25\,A
\end{aligned}
\]
(C) Potential difference across lamp and Conductor
Let Potential Difference across lamp=\(V_l\) and across Conductor \(V_c\) and current is \(I\)
\[\begin{aligned}
V_l&=IR_l\\
&=0.25\times 20 \\
&=5\,V\\\\
V_c&=I\times R_c\\
&=0.25\times 4\\
&=1\,V
\end{aligned}\]
Example-10
In the circuit diagram given in Fig. 11.10, suppose the resistors
\(R_1, R_2,\text{ and }R_3\)
have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have
been connected to a battery of 12 V.
Calculate
(a) the current through
each resistor,
(b) the total current in the circuit, and
(c) the total circuit
resistance.
Solution
Resistances \(R_1=5/,\Omega\)\(R_2=10/,\Omega\)
\(R_3=30/,\Omega\)
Voltage = 12 V
(a) the current through each resistor Let Current through Resistor \(R_1, R_2,\text{ and }R_3\) be \(I_1,\,I_2\text{ and }I_3\)
\[ \begin{aligned} I_1&=\frac{V}{R_1}\\ &=\frac{12}{5}\\ &=2.4\,\Omega\\\\ I_2&=\frac{V}{R_2}\\ &=\frac{12}{10}\\ &=1.2\,\Omega\\\\ I_3&=\frac{V}{R_3}\\ &=\frac{12}{30}\\ &=0.4\,\Omega \end{aligned} \] (b) the total current in the circuit
Let Total Current in the circuit be \(I\) Total Resistance \(R_p\) \[ \begin{aligned} \frac{1}{R_p}&=\frac{1}{5} + \frac{1}{10}+ \frac{1}{30}\\\\ &=\frac{6+3+1}{30}\\\\ &=\frac{10}{30}\\\\ &=\frac{1}{3}\\\\ \implies R_p&=3,\,\Omega \end{aligned} \] \[ \begin{aligned} I&=\frac{V}{R_p}\\\\ &=\frac{12}{3}\\\\ &=4\,A \end{aligned} \] (c) the total circuit resistance. = \(R_p=3\,\Omega\)
Example-11
If in Fig. 11.12,
R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω,
and a 12 V battery is connected to the arrangement. Calculate
(a) the total resistance in the circuit, and
(b) the total current flowing
in the circuit.
Solution
\[ \text{Resistances as given:} \quad \begin{cases} R_{1} = 10\,\Omega \\[4pt] R_{2} = 40\,\Omega \\[4pt] R_{3} = 30\,\Omega \\[4pt] R_{4} = 20\,\Omega \\[4pt] R_{5} = 60\,\Omega \end{cases} \]
\(R_1\), and \(R_2\), are in Parallel Arrangement
Let \(R_u\) is the equivalent Resistance of \(R_1\) ,& \(R_2\)
\(R_3,\, R_4, \text{ and }R_3\) are also in parallel Arrangement
Let \(R_l\) is the equivalent Resistance of \(R_3,\, R_4, \text{ and }R_3\)
let R be the equivalent Resistance of the \(Ru\) & \(R_l\) $$\begin{aligned}R&=R_{u}+R_{l}\\ &=8+10\\ &=18\Omega \end{aligned}$$
Thus, Total Resistance of the circuit is \(=18\,\Omega\)
Voltage = 12 V Resistance = 18 \(\Omega\)
$$\begin{aligned}I&=\dfrac{V}{R}\\ &=\dfrac{12}{18}\\ &\approx 0.67\,A\end{aligned}$$Thus, Total current in the circuit is \(\approx 0.67\,A\)
HEATING EFFECT OF ELECTRIC CURRENT
When an electric current passes through a conducting wire, the electrical energy supplied by the
source
does not remain unchanged. A part of this energy is transformed into heat. This phenomenon is known
as the
Heating Effect of Electric Current.
Consider a current \(I\) flowing through a resistor of resistance \(R\). Let
the potential difference across it be \(V\) (Fig. 11.13). Let \(t\) be the time during
which a charge \(Q\) flows across. The work done in moving the charge \(Q\)
through a potential difference \(V\) is \(VQ\). Therefore, the source must supply
energy equal to \(VQ\) in time \(t\). Hence the power input to the circuit by the
source is
Or the energy supplied to the circuit by the source in time \(t\) is \(P \times t\),
that is, \(VIt\).
This energy gets dissipated in the resistor as heat. Thus for a steady
current \(I\), the amount of heat \(H\) produced in time \(t\) is
\[\boxed{H=I^2Rt}\]
Example-12
An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
Solution
Max. Energy consumed =480WVoltage =220V \[\begin{aligned}I&=\frac{P}{V}\\\\ &=\frac{840}{220}\\\\ &=3.82\,A\\\\ R&=\frac{V}{I}\\\\ &=\frac{220}{3.82}\\\\ &=57.60\,\Omega\end{aligned}\] Min Energy=360 Max. Energy consumed =480W
Voltage =220V \[\begin{aligned}I&=\frac{P}{V}\\\\ &=\frac{360}{220}\\\\ &\approx 1.64\,A\\\\ R&=\frac{V}{I}\\\\ &=\frac{220}{1.64}\\\\ &\approx134.15\,\Omega\end{aligned}\]
Example-13
100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.
Solution
Heat Produced =100J
Time= 1 s
Resistance (R) =\(4\Omega\)
Let Potential Difference be V
Potential across Resistance when
I=5 A
R=4 \(\Omega\)
Practical Applications of Heating Effect of Electric Current
Common real-world devices that utilise the heating effect of an electric current.
Electric Iron
A high-resistance coil (usually nichrome) inside the iron gets heated when current flows. The generated heat warms the soleplate to press and smooth clothes.
- Uses a thermostatic control to regulate temperature
- High resistance ⇒ greater heat generation
Electric Heater / Geyser
Heating elements convert electrical energy to heat for warming air or water. Elements are designed with suitable resistance and safety cut-offs.
- Common in room heaters, immersion rods, and water geysers
- Often include safety features like thermostats and fuses
Electric Bulb (Incandescent)
The tungsten filament has high resistance; when heated by current it glows and emits light (and a significant amount of heat).
- Not energy-efficient compared to modern LEDs
- Illustrates simultaneous production of light and heat
Electric Fuse
A thin fusible wire is designed to melt when excessive current produces dangerous heat — protecting wiring and appliances from damage.
- Acts as a safety device against overcurrent
- Melting point and wire thickness are chosen carefully
Industrial & Other Uses
Heating effect is applied in welding, metal cutting, soldering, and electric furnaces where controlled high temperatures are required.
- Electric furnaces for melting and heat treatment
- Electrical heating elements in kitchens and laboratories
ELECTRIC POWER
Electric power is defined as the rate of doing electrical work or the rate of consumption of electrical energy in a circuit.
\[\boxed{P=VI}\] or, \[\boxed{P=\frac{V^2}{R}}\]
The SI unit of power is:
Watt (W)
One watt is the power consumed by a device that uses 1 joule of energy in 1 second.
Example-14
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Solution
Voltage V =220
Current=0.50 A
\[\begin{aligned}
P&=VI\\
&=220\times 0.50\\
&=110\,W\end{aligned}\]
Example-15
An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?
Solution
Power of generator =400W
Operation Time =8 hours/day
Energy Consumed in one day
\[\begin{aligned}P&=\frac{W}{t}\\
\implies W&=P\times t\\
&=400\times 8\times\\
&=3200\,Wh\\
&=3.2\,kWh\end{aligned}\]
Therefore, Energy consumed in 30 days
\[
\begin{aligned}
E&=3.2\times 30\\
&=96kWh
\end{aligned}
\]
Therfore, Cost of Electricity=
\[\begin{aligned}
96\times 3\\
=\text{₹ }288.00\end{aligned}\]
Important Points
- A stream of electrons moving through a conductor constitutes an electric current. Conventionally, the direction of current is taken opposite to the direction of flow of electrons.
- The SI unit of electric current is ampere.
- To set the electrons in motion in an electric circuit, we use a cell or a battery. A cell generates a potential difference across its terminals. It is measured in volts (V).Resistance is a property that resists the flow of electrons in a conductor. It controls the magnitude of the current. The SI unit of resistance is ohm (Ω).
- Ohm’s law: The potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same.
- The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor.
- The equivalent resistance of several resistors in series is equal to the sum of their individual resistances.
- A set of resistors connected in parallel has an equivalent resistance \(R_p \) given by \[\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots\]
- The electrical energy dissipated in a resistor is given by \(W = V \times I \times t\)
- The unit of power is watt (W). One watt of power is consumed when 1 A of current flows at a potential difference of 1 V.
- The commercial unit of electrical energy is kilowatt hour (kWh). 1 kW h = 3,600,000 J = 3.6 × 106 J.
