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Current Electricity Class 12 Physics Notes

Introduction

While previous studies focused on charges at rest (static electricity), this chapter explores electric current, which is the phenomenon of charges in motion.

You see electric currents in nature and in daily life. Lightning is a dramatic, unsteady flow of charge from clouds to the earth. In contrast, devices like a flashlight or a battery-powered clock rely on a steady, continuous flow of charge, much like water flowing smoothly in a river. We will focus on the basic laws that govern these steady electric currents.

Electric Current

Imagine an area positioned perpendicular to the flow of electric charges. Both positive and negative charges can move across this area.

The electric current is defined as the net rate of flow of charge across an area.

If a net charge $q$ flows across an area in time $t$ , the steady current $I$ is given by the formula: $I = \frac{q}{t}$ Here, the net charge $q$ is the amount of positive charge moving forward ( $q_+$ ) minus the amount of negative charge moving forward ( $q_-$ ). So, $q = q_+ - q_-$ . A negative value for $I$ simply means the net current is flowing in the backward direction.
Since current is not always steady, a more general definition is used for the instantaneous current at a specific time $t$ . If a small amount of charge $\Delta Q$ flows across a cross-section in a small time interval $\Delta t$ , the current is: $I(t) = \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t}$

The SI unit of current is the ampere (A). [!example] Currents in household appliances are typically in the range of a few amperes. In contrast, an average lightning strike can carry tens of thousands of amperes, while the electrical signals in our nerves involve currents of only a few microamperes ( $10^{-6}$ A).

Electric Currents in Conductors

An electric charge will experience a force and move if it is placed in an electric field, creating a current.

In some materials, like metals, some electrons are "free" to move throughout the material. These materials are called conductors.
In other materials, called insulators, electrons are tightly bound to their atoms and cannot move freely, even when an electric field is applied.

In solid conductors like copper wires, the current is carried by the movement of negatively charged electrons, while the positive atomic nuclei (ions) remain fixed in place. In other types of conductors, such as electrolytic solutions (saltwater), both positive and negative ions can move and contribute to the current.

Electron Motion in a Conductor

Without an Electric Field: The free electrons in a metal are in constant, random motion due to thermal energy. They frequently collide with the fixed ions. After each collision, an electron's direction is completely random. On average, for every electron moving in one direction, another is moving in the opposite direction. Therefore, there is no net flow of charge, and thus no electric current.
With an Electric Field: When an electric field is applied across a conductor (for example, by connecting it to a battery), the free electrons are accelerated by the field. They begin to drift in a direction opposite to the electric field (towards the positive potential). This collective, directional movement of electrons constitutes an electric current. To maintain a steady current, a device like a cell or battery is needed to continuously supply energy and maintain the electric field.

Ohm's Law

In 1828, G.S. Ohm discovered a fundamental relationship concerning the flow of current.

Ohm's Law states that the potential difference ( $V$ ) across the ends of a conductor is directly proportional to the current ( $I$ ) flowing through it, provided the temperature and other physical conditions remain unchanged.

Mathematically, this is expressed as: $V \propto I$ or $V = RI$

The constant of proportionality, $R$ , is called the resistance of the conductor.

Resistance (R) is a measure of how much a material opposes the flow of electric current.
The SI unit of resistance is the ohm, denoted by the symbol $\Omega$ . $1 \Omega = 1 \text{ V A}^{-1}$ .

Factors Affecting Resistance

The resistance of a conductor depends on its material and its physical dimensions (length and area).

Length ( $l$ ): Resistance is directly proportional to the length of the conductor. A longer wire has more resistance. $R \propto l$
Cross-sectional Area ( $A$ ): Resistance is inversely proportional to the cross-sectional area of the conductor. A thicker wire has less resistance. $R \propto \frac{1}{A}$

Combining these relationships, we get: $R = \rho \frac{l}{A}$ The constant $\rho$ (rho) is called the resistivity (or specific resistance) of the material. Resistivity is an intrinsic property of a material, meaning it does not depend on the conductor's shape or size, but on the nature of the material itself.

Current Density and Vector Form of Ohm's Law

Current density (j) is defined as the current per unit area normal to the direction of current flow. $j = \frac{I}{A}$ The SI unit for current density is amperes per square meter ( $\text{A/m}^2$ ).

Using the relationship between electric field ( $E$ ) and potential difference ( $V$ ) for a uniform field over a length $l$ , $V = El$ , we can express Ohm's law in another form: $E = j\rho$ Since the direction of current flow (for positive charges) is the same as the direction of the electric field, this can be written in vector form: $\mathbf{E} = \rho\mathbf{j}$ or $\mathbf{j} = \sigma\mathbf{E}$ Here, $\sigma$ (sigma) is the conductivity of the material, which is simply the reciprocal of resistivity ( $\sigma = 1/\rho$ ).

Drift of Electrons and the Origin of Resistivity

The microscopic behavior of electrons in a conductor explains why materials have resistance and obey Ohm's law.

When an electric field $\mathbf{E}$ is applied, each free electron (charge $-e$ , mass $m$ ) experiences a force and accelerates: $\mathbf{a} = \frac{-e\mathbf{E}}{m}$ However, electrons do not accelerate indefinitely. Their path is interrupted by frequent collisions with the fixed positive ions of the metal. After each collision, the electron's velocity is randomized.

Between collisions, the electron gains a small velocity component in the direction opposite to the electric field. The average velocity of all electrons due to this effect is called the drift velocity ( $\mathbf{v}_d$ ).

The drift velocity is constant for a steady electric field and is given by: $\mathbf{v}_d = -\frac{e\mathbf{E}}{m}\tau$ Here, $\tau$ (tau) is the relaxation time, which is the average time between successive collisions for an electron.

Relating Drift Velocity to Current and Resistivity

The total current $I$ flowing through a conductor with cross-sectional area $A$ is related to the drift velocity by: $I = neA|v_d|$ where $n$ is the number of free electrons per unit volume (electron number density).

By substituting the expression for drift velocity into this equation, we can derive a microscopic version of Ohm's law. The conductivity $\sigma$ and resistivity $\rho$ can be expressed in terms of microscopic quantities:

Conductivity: $\sigma = \frac{ne^2\tau}{m}$
Resistivity: $\rho = \frac{1}{\sigma} = \frac{m}{ne^2\tau}$

This model shows that a material's resistivity depends on the number of free charge carriers ( $n$ ) and the average time between their collisions ( $\tau$ ).

Example

(a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area

1.0 \times 10^{-7} \text{ m}^2

carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is

9.0 \times 10^3 \text{ kg/m}^3

, and its atomic mass is 63.5 u. (b) Compare the drift speed obtained above with, (i) thermal speeds of copper atoms at ordinary temperatures, (ii) speed of propagation of electric field along the conductor which causes the drift motion.

Given

Cross-sectional area, $A = 1.0 \times 10^{-7} \text{ m}^2$
Current, $I = 1.5 \text{ A}$
Density of copper, $\rho_{Cu} = 9.0 \times 10^3 \text{ kg/m}^3$
Atomic mass of copper = 63.5 u (or 63.5 g/mol)
Avogadro's number, $N_A = 6.0 \times 10^{23} \text{ atoms/mol}$
Charge of an electron, $e = 1.6 \times 10^{-19} \text{ C}$

To Find

(a) Drift speed, $v_d$ (b) Comparison of $v_d$ with (i) thermal speed and (ii) speed of electric field.

Formula

$v_d = \frac{I}{neA}$ Number of atoms per cubic meter, $n = \frac{\text{Avogadro's number}}{\text{Molar mass}} \times \text{Density}$

Solution

(a) Calculate the drift speed

First, we need to find the number density of conduction electrons, $n$ . Since each copper atom contributes one electron, $n$ is equal to the number of atoms per cubic metre. A cubic metre of copper has a mass of $9.0 \times 10^3 \text{ kg} = 9.0 \times 10^6 \text{ g}$ . The number of moles in this mass is $\frac{9.0 \times 10^6 \text{ g}}{63.5 \text{ g/mol}}$ . So, the number of atoms is: $n = \frac{6.0 \times 10^{23} \text{ atoms/mol}}{63.5 \text{ g/mol}} \times 9.0 \times 10^6 \text{ g/m}^3$ $n = 8.5 \times 10^{28} \text{ m}^{-3}$ Now, we can calculate the drift speed: $v_d = \frac{1.5 \text{ A}}{(8.5 \times 10^{28} \text{ m}^{-3}) \times (1.6 \times 10^{-19} \text{ C}) \times (1.0 \times 10^{-7} \text{ m}^2)}$ $v_d = 1.1 \times 10^{-3} \text{ m s}^{-1} = 1.1 \text{ mm s}^{-1}$

Answer for part (a) = $1.1 \text{ mm s}^{-1}$

(b) Comparison of speeds

(i) The typical thermal speed of copper atoms at ordinary temperatures (300 K) is about $2 \times 10^2 \text{ m/s}$ . The electron drift speed ( $1.1 \times 10^{-3} \text{ m/s}$ ) is incredibly slow in comparison, about $10^{-5}$ times smaller.

(ii) An electric field propagates along a conductor at the speed of an electromagnetic wave, which is the speed of light, $3.0 \times 10^8 \text{ m s}^{-1}$ . The drift speed is extremely small compared to this, smaller by a factor of about $10^{-11}$ .

Note

Even though the drift speed of electrons is very slow (a few millimeters per second), a current is established almost instantly when a circuit is closed. This is because the electric field travels through the wire at nearly the speed of light, causing all the free electrons throughout the wire to start drifting almost simultaneously.

Example

(a) In Example 3.1, the electron drift speed is estimated to be only a few

\text{mm s}^{-1}

for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed? (b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed? (c) If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor? (d) When electrons drift in a metal from lower to higher potential, does it mean that all the 'free' electrons of the metal are moving in the same direction? (e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field?

Solution

(a) An electric field is established throughout the circuit almost instantly (at the speed of light), causing a local electron drift at every point. The establishment of a current does not require electrons to travel from one end of the conductor to the other.

(b) Each free electron does accelerate due to the electric field, but this acceleration is constantly interrupted by collisions with the positive ions of the metal. After each collision, the electron loses its drift speed and must accelerate again. This repeated process of acceleration and collision results in a constant, steady average drift speed.

(c) We can obtain large currents because the number density of free electrons in a conductor is enormous, on the order of $10^{29} \text{ m}^{-3}$ . Even with a tiny drift speed, the sheer number of moving charges results in a significant current.

(d) Not at all. The very slow drift velocity is superimposed on the much larger, random thermal velocities of the electrons. The electrons are still moving randomly in all directions at high speeds, but with a tiny, collective "drift" in one direction.

(e) (i) In the absence of an electric field, the paths of electrons between collisions are straight lines. (ii) In the presence of an electric field, the force on the electrons causes their paths between collisions to be curved.

Mobility

Mobility ( $\mu$ ) is a quantity that describes how quickly a charge carrier (like an electron) can move through a material when an electric field is applied. It is defined as the magnitude of the drift velocity per unit electric field: $\mu = \frac{|\mathbf{v}_d|}{E}$ The SI unit of mobility is $\text{m}^2/(\text{V} \cdot \text{s})$ or $\text{m}^2 \text{V}^{-1} \text{s}^{-1}$ .

Using the formula for drift velocity, mobility can also be expressed in terms of the relaxation time: $\mu = \frac{e\tau}{m}$

Limitations of Ohm's Law

While Ohm's law is valid for a wide range of materials (called ohmic materials), it is not a fundamental law of nature. Many materials and devices, especially in electronics, do not obey Ohm's law (they are non-ohmic). Deviations from Ohm's law typically fall into one of the following categories:

Non-linear V-I relationship: The potential difference $V$ ceases to be directly proportional to the current $I$ . For example, in a good conductor, as the current increases, it heats up, increasing its resistance. This causes the V-I graph to curve.
Dependence on the sign of V: The relationship between $V$ and $I$ depends on the direction of the voltage. A diode, for example, allows current to flow easily in one direction (forward bias) but offers very high resistance in the opposite direction (reverse bias).
Non-unique V-I relationship: The same current $I$ can correspond to more than one value of voltage $V$ . A material called Gallium Arsenide (GaAs) exhibits this behavior.

Resistivity of Various Materials

Materials can be classified based on their resistivity ( $\rho$ ):

Conductors: Have very low resistivities, typically in the range of $10^{-8} \Omega\text{m}$ to $10^{-6} \Omega\text{m}$ . Metals are excellent conductors.
Insulators: Have very high resistivities, such as ceramic, rubber, and plastics. Their resistivities can be $10^{18}$ times greater than that of metals.
Semiconductors: Have resistivities that fall between those of conductors and insulators. Materials like silicon (Si) and germanium (Ge) are semiconductors. Their resistivity can be precisely controlled by adding impurities, a feature that is crucial for making electronic devices.

Temperature Dependence of Resistivity

The resistivity of a material changes with temperature.

Metals: For most metals, resistivity increases as temperature increases. Over a limited temperature range, this relationship is approximately linear and can be described by the formula: $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$ where $\rho_T$ is the resistivity at temperature $T$ , $\rho_0$ is the resistivity at a reference temperature $T_0$ , and $\alpha$ is the temperature coefficient of resistivity. For metals, $\alpha$ is positive.
Alloys: Some alloys, like Nichrome, Manganin, and Constantan, have resistivities that change very little with temperature. This property makes them ideal for manufacturing standard resistors used in measurement instruments.
Semiconductors and Insulators: For these materials, resistivity decreases as temperature increases.

Microscopic Explanation

We can understand this temperature dependence using the formula $\rho = \frac{m}{ne^2\tau}$ .

In a metal, the number density of free electrons ( $n$ ) is nearly constant. As temperature rises, the ions vibrate more vigorously, leading to more frequent collisions with electrons. This reduces the relaxation time ( $\tau$ ), and since $\rho$ is inversely proportional to $\tau$ , the resistivity increases.
In a semiconductor, increasing the temperature provides enough energy to free more electrons, significantly increasing the number density of charge carriers ( $n$ ). This increase in $n$ is far more impactful than the decrease in $\tau$ . Since $\rho$ is inversely proportional to $n$ , the resistivity decreases.

Example

An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (

27.0^\circ\text{C}

) is found to be

75.3 \Omega

. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is

1.70 \times 10^{-4} {}^\circ\text{C}^{-1}

.

Given

Initial temperature (room temp), $T_1 = 27.0^\circ\text{C}$
Initial resistance, $R_1 = 75.3 \Omega$
Supply voltage, $V = 230 \text{ V}$
Steady current, $I_2 = 2.68 \text{ A}$
Temperature coefficient of resistance, $\alpha = 1.70 \times 10^{-4} {}^\circ\text{C}^{-1}$

To Find

The steady temperature of the element, $T_2$ .

Formula

Ohm's Law: $R_2 = \frac{V}{I_2}$
Temperature dependence of resistance: $R_2 = R_1[1 + \alpha(T_2 - T_1)]$

Solution

First, calculate the resistance of the element at its steady operating temperature, $R_2$ . $R_2 = \frac{230 \text{ V}}{2.68 \text{ A}} = 85.8 \Omega$ Now, use the temperature dependence formula to find the change in temperature, $(T_2 - T_1)$ . $R_2 = R_1[1 + \alpha(T_2 - T_1)]$ Rearranging for $(T_2 - T_1)$ : $T_2 - T_1 = \frac{R_2 - R_1}{R_1 \alpha}$ $T_2 - T_1 = \frac{(85.8 - 75.3) \Omega}{75.3 \Omega \times 1.70 \times 10^{-4} {}^\circ\text{C}^{-1}} = \frac{10.5}{0.012801} {}^\circ\text{C} \approx 820^\circ\text{C}$ Now, find the final temperature $T_2$ : $T_2 = T_1 + 820^\circ\text{C} = 27.0^\circ\text{C} + 820^\circ\text{C} = 847^\circ\text{C}$

Final Answer The steady temperature of the nichrome element is $847^\circ\text{C}$ .

Example

The resistance of the platinum wire of a platinum resistance thermometer at the ice point (

0^\circ\text{C}

) is

5 \Omega

and at steam point (

100^\circ\text{C}

) is

5.23 \Omega

. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is

5.795 \Omega

. Calculate the temperature of the bath.

Given

Resistance at ice point, $R_0 = 5 \Omega$
Resistance at steam point, $R_{100} = 5.23 \Omega$
Resistance at unknown temperature $t$ , $R_t = 5.795 \Omega$

To Find

The temperature of the bath, $t$ .

Formula

Assuming a linear relationship between resistance and temperature, the temperature $t$ can be found using the formula: $t = \frac{R_t - R_0}{R_{100} - R_0} \times 100^\circ\text{C}$

Solution

Substitute the given values into the formula: $t = \frac{5.795 - 5}{5.23 - 5} \times 100$ $t = \frac{0.795}{0.23} \times 100 = 345.65^\circ\text{C}$

Final Answer The temperature of the bath is $345.65^\circ\text{C}$ .

Electrical Energy, Power

When a current $I$ flows through a conductor with a potential difference $V$ across it, the electric field does work on the charges. In a time interval $\Delta t$ , a charge $\Delta Q = I\Delta t$ moves through the conductor. The change in potential energy is: $\Delta U_{pot} = -\Delta Q V = -IV\Delta t$ This loss in potential energy is not converted into kinetic energy (since the electrons move at a constant drift velocity). Instead, due to collisions with the lattice ions, the energy is converted into heat.

The amount of energy dissipated as heat, $\Delta W$ , in time $\Delta t$ is: $\Delta W = IV\Delta t$

Electric power (P) is the rate at which this energy is dissipated. $P = \frac{\Delta W}{\Delta t} = IV$ Using Ohm's law ( $V=IR$ ), we can express power in two other useful forms: $P = I^2R$ $P = \frac{V^2}{R}$ This power dissipation is often called ohmic loss or Joule heating. It is the principle behind electric heaters, toasters, and incandescent light bulbs.

Power Transmission

Electrical power is generated at power stations and transmitted over long distances through cables. These cables have resistance ( $R_c$ ), which leads to power loss. The power wasted in the cables is given by $P_c = I^2R_c$ .

To deliver a certain amount of power $P = VI$ to consumers, the current is $I = P/V$ . The power loss in the cables is therefore: $P_c = \left(\frac{P}{V}\right)^2 R_c = \frac{P^2R_c}{V^2}$ This equation shows that the power loss is inversely proportional to the square of the transmission voltage ( $V^2$ ). To minimize this loss, electricity is transmitted at extremely high voltages. A transformer is then used to step down the voltage to a safer level for use in homes and factories.

Cells, emf, Internal Resistance

An electrolytic cell is a device that maintains a steady current in a circuit. It consists of two electrodes, a positive (P) and a negative (N), immersed in an electrolyte.

Electromotive Force (emf, $\varepsilon$ ): Due to chemical reactions, a potential difference is created between the electrodes and the electrolyte. The emf of a cell is defined as the potential difference between its positive and negative terminals when no current is flowing (i.e., in an open circuit). Emf is a voltage, not a force, and represents the work done per unit charge by the cell.
Internal Resistance (r): The electrolyte itself has some resistance to the flow of current. This is called the internal resistance of the cell.

When a cell of emf $\varepsilon$ and internal resistance $r$ is connected to an external resistor $R$ , a current $I$ flows. The potential difference across the external resistor, known as the terminal voltage (V), is less than the emf because of the voltage drop across the internal resistance. $V = \varepsilon - Ir$ Since $V=IR$ for the external resistor, we can write: $IR = \varepsilon - Ir$ Solving for the current $I$ in the circuit gives: $I = \frac{\varepsilon}{R+r}$ The maximum current a cell can deliver is when the external resistance is zero ( $R=0$ ), which is $I_{max} = \varepsilon/r$ .

Cells in Series and in Parallel

Cells can be combined to achieve a desired voltage or current capacity.

Cells in Series

When cells are connected in series, the negative terminal of one cell is connected to the positive terminal of the next. For two cells with emfs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ :

The equivalent emf is the sum of the individual emfs: $\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$
The equivalent internal resistance is the sum of the individual internal resistances: $r_{eq} = r_1 + r_2$ For $n$ cells in series, $\varepsilon_{eq} = \sum \varepsilon_i$ and $r_{eq} = \sum r_i$ .

Note

If a cell is connected with its polarity reversed (positive to positive), its emf is subtracted from the total. For example, if cell 2 is reversed,

\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2

.

Cells in Parallel

When cells are connected in parallel, all positive terminals are connected to one point and all negative terminals to another. For two cells:

The equivalent internal resistance is given by: $\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \quad \text{or} \quad r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$
The equivalent emf is given by: $\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \quad \text{or} \quad \varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$ These formulas can be extended for any number of cells in parallel.

Kirchhoff's Rules

For complex circuits that cannot be simplified using series and parallel rules, we use two rules developed by Gustav Kirchhoff.

Junction Rule (Kirchhoff's First Rule): At any junction (a point where multiple wires meet), the sum of the currents entering the junction must equal the sum of the currents leaving it. $\sum I_{in} = \sum I_{out}$ This rule is a statement of the conservation of charge. Since charge cannot accumulate at a junction, the rate at which it flows in must equal the rate at which it flows out.
Loop Rule (Kirchhoff's Second Rule): The algebraic sum of the changes in potential around any closed loop in a circuit must be zero. $\sum \Delta V_{loop} = 0$ This rule is a statement of the conservation of energy. If you start at any point in a circuit and travel around a closed loop back to the starting point, your electric potential must be the same as when you started.

Example

A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance

1 \Omega

(Fig. 3.16). Determine the equivalent resistance of the network and the current along each edge of the cube.

Given

Emf of battery, $\varepsilon = 10 \text{ V}$
Resistance of each edge, $R = 1 \Omega$
The circuit is a cubical network of 12 identical resistors.

To Find

Equivalent resistance of the network, $R_{eq}$
Current along each edge.

Formula

Kirchhoff's Loop Rule
Ohm's Law

Solution

Let the total current from the battery be $3I$ . Due to the symmetry of the cube, this current splits equally into the three paths from corner A (AB, AD, AA'). So, the current in each of these edges is $I$ .

At corners B, D, and A', the current $I$ again splits equally into two paths. So, the current in edges like BC, B'B, DC, etc., is $I/2$ .

At corners C, C', and D', currents from two edges (each $I/2$ ) combine. So, the current in edges like C'C, D'C, etc., is $I$ .

Finally, at corner C', the three currents of value $I$ combine to form the total current $3I$ , which returns to the battery.

Now, apply Kirchhoff's loop rule to a closed loop, for example, ABCC'EA. (E represents the battery). The potential changes are:

Across AB: $-IR$
Across BC: $-(I/2)R$
Across CC': $-IR$
Across the battery: $+\varepsilon$

The sum of potential changes is zero: $+\varepsilon - IR - (I/2)R - IR = 0$ $\varepsilon = IR + (I/2)R + IR = \frac{5}{2}IR$ The total current flowing from the source is $3I$ . The equivalent resistance $R_{eq}$ of the network is given by Ohm's law for the entire circuit: $R_{eq} = \frac{\varepsilon}{\text{Total Current}} = \frac{\varepsilon}{3I}$ Substitute the expression for $\varepsilon$ : $R_{eq} = \frac{(5/2)IR}{3I} = \frac{5}{6}R$ For $R=1 \Omega$ , the equivalent resistance is: $R_{eq} = \frac{5}{6} \Omega$ The total current is: $\text{Total Current} = 3I = \frac{\varepsilon}{R_{eq}} = \frac{10 \text{ V}}{(5/6) \Omega} = 12 \text{ A}$ From this, we find $I$ : $I = \frac{12 \text{ A}}{3} = 4 \text{ A}$ The currents in the edges are:

Edges like AB, AD, AA': $I = 4 \text{ A}$
Edges like BC, CD, A'B': $I/2 = 2 \text{ A}$
Edges like C'C, B'C', D'C': $I = 4 \text{ A}$

Final Answer The equivalent resistance is $(5/6) \Omega$ . The currents are 4 A in the edges nearest the input/output corners, and 2 A in the remaining edges.

Example

Determine the current in each branch of the network shown in Fig. 3.17.

Given

A network of resistors and cells as shown in the figure.

Resistors: $4 \Omega, 2 \Omega, 1 \Omega, 4 \Omega, 2 \Omega$
Cells: $10 \text{ V}, 5 \text{ V}$

To Find

The current in each branch of the network.

Formula

Kirchhoff's Junction Rule
Kirchhoff's Loop Rule

Solution

Let's assign currents to the branches. Let the current from the 10V cell be $I_1$ . At junction A, let the current in branch AB be $I_2$ . By the junction rule, the current in branch AD is $(I_1 - I_2)$ . At junction B, let the current in branch BC be $I_3$ . Then the current in branch BD is $(I_2 + I_3)$ . At junction C, currents $I_1$ and $I_3$ meet, so the current in the 10V cell is $(I_1+I_3)$ . This contradicts our initial assumption. A better way is to assign separate variables and use junction rules.

Let's use the current distribution given in the source text's solution for clarity.

Current in branch CA: $I_1$
Current in branch AB: $I_2$
Current in branch BC: $I_1$ (from C to B)
Current in branch AD: $I_1 - I_2$
Current in branch CD: $I_2 + I_3 - I_1$
Current in branch DEB: $I_3$

Apply Kirchhoff's loop rule to three closed loops:

1. Loop ADCA: Start at A and go clockwise: A -> D -> C -> A

Across AD (4 $\Omega$ ): $-4(I_1 - I_2)$
Across DC (2 $\Omega$ ): $+2(I_2 + I_3 - I_1)$
Across CA (1 $\Omega$ ): $-1(I_1)$
Across cell (10V): $+10$ $10 - 4(I_1 - I_2) + 2(I_2 + I_3 - I_1) - I_1 = 0$ $10 - 4I_1 + 4I_2 + 2I_2 + 2I_3 - 2I_1 - I_1 = 0$ $7I_1 - 6I_2 - 2I_3 = 10 \quad \text{(Eq. 1)}$

2. Loop ABCA: Start at A and go clockwise: A -> B -> C -> A

Across AB (4 $\Omega$ ): $-4I_2$
Across BC (2 $\Omega$ ): $-2(I_2 + I_3)$
Across CA (1 $\Omega$ ): $-1(I_1)$
Across cell (10V): $+10$ $10 - 4I_2 - 2(I_2 + I_3) - I_1 = 0$ $10 - 4I_2 - 2I_2 - 2I_3 - I_1 = 0$ $I_1 + 6I_2 + 2I_3 = 10 \quad \text{(Eq. 2)}$

3. Loop BCDEB: Start at B and go clockwise: B -> C -> D -> E -> B

Across BC (2 $\Omega$ ): $-2(I_2 + I_3)$
Across CD (2 $\Omega$ ): $-2(I_2 + I_3 - I_1)$
Across cell (5V): $+5$ $5 - 2(I_2 + I_3) - 2(I_2 + I_3 - I_1) = 0$ $5 - 2I_2 - 2I_3 - 2I_2 - 2I_3 + 2I_1 = 0$ $2I_1 - 4I_2 - 4I_3 = -5 \quad \text{(Eq. 3)}$

Solving these three simultaneous equations gives: $I_1 = 2.5 \text{ A}$ $I_2 = \frac{5}{8} \text{ A} = 0.625 \text{ A}$ $I_3 = 1 \frac{7}{8} \text{ A} = 1.875 \text{ A}$

Now we find the current in each branch:

Branch AB: $I_2 = \frac{5}{8} \text{ A}$
Branch CA: $I_1 = 2.5 \text{ A}$
Branch DEB: $I_3 = 1 \frac{7}{8} \text{ A}$
Branch AD: $I_1 - I_2 = 2.5 - 0.625 = 1.875 \text{ A} = 1 \frac{7}{8} \text{ A}$
Branch CD: $I_2 + I_3 - I_1 = 0.625 + 1.875 - 2.5 = 0 \text{ A}$
Branch BC: $I_2 + I_3 = 0.625 + 1.875 = 2.5 \text{ A}$

Final Answer The currents are: AB: $5/8$ A, CA: $2.5$ A, DEB: $1.875$ A, AD: $1.875$ A, CD: $0$ A, BC: $2.5$ A.

Wheatstone Bridge

The Wheatstone bridge is a circuit used to accurately measure an unknown electrical resistance. It consists of four resistors ( $R_1, R_2, R_3, R_4$ ) arranged in a diamond shape.

A voltage source (battery) is connected across one pair of opposite corners (the battery arm).
A sensitive current detector, called a galvanometer (G), is connected across the other pair of opposite corners (the galvanometer arm).

The bridge is said to be balanced when the resistors are adjusted so that no current flows through the galvanometer ( $I_g = 0$ ). When the bridge is balanced, the potential at both ends of the galvanometer arm is the same.

By applying Kirchhoff's rules to a balanced bridge, we arrive at the balance condition: $\frac{R_2}{R_1} = \frac{R_4}{R_3}$ This can be rearranged as: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$

To use the bridge, you would place an unknown resistor (say, $R_4$ ) in one arm. Then, you would adjust a variable resistor (say, $R_3$ ) until the galvanometer reads zero. At that point, the bridge is balanced, and the unknown resistance can be calculated if the other three resistances are known: $R_4 = R_3 \frac{R_2}{R_1}$

Example

The four arms of a Wheatstone bridge (Fig. 3.19) have the following resistances: AB =

100 \Omega

, BC =

10 \Omega

, CD =

5 \Omega

, and DA =

60 \Omega

. A galvanometer of

15 \Omega

resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Given

Resistance AB, $R_{AB} = 100 \Omega$
Resistance BC, $R_{BC} = 10 \Omega$
Resistance CD, $R_{CD} = 5 \Omega$
Resistance DA, $R_{DA} = 60 \Omega$
Resistance of galvanometer, $R_g = 15 \Omega$
Potential difference across AC, $V_{AC} = 10 \text{ V}$

To Find

The current through the galvanometer, $I_g$ .

Formula

Kirchhoff's Loop Rule

Solution

Let's assign currents:

Current in branch DA is $I_2$
Current in branch AB is $I_1$
Current in branch BD (galvanometer) is $I_g$ Using the junction rule at B and D:
Current in branch BC is $I_1 - I_g$
Current in branch DC is $I_2 + I_g$

Apply Kirchhoff's loop rule to three loops:

1. Loop BADB: $-100I_1 - 15I_g + 60I_2 = 0$ Dividing by -5: $20I_1 + 3I_g - 12I_2 = 0 \quad \text{(Eq. 1)}$

2. Loop BCDB: $-10(I_1 - I_g) + 5(I_2 + I_g) + 15I_g = 0$ $-10I_1 + 10I_g + 5I_2 + 5I_g + 15I_g = 0$ $-10I_1 + 30I_g + 5I_2 = 0$ Dividing by -5: $2I_1 - 6I_g - I_2 = 0 \quad \text{(Eq. 2)}$

3. Loop ADCEA (through the source): $-60I_2 - 5(I_2 + I_g) + 10 = 0$ $-60I_2 - 5I_2 - 5I_g + 10 = 0$ $-65I_2 - 5I_g + 10 = 0$ Dividing by -5: $13I_2 + I_g = 2 \quad \text{(Eq. 3)}$

Now we solve these three simultaneous equations. From Eq. 2, express $I_2$ in terms of $I_1$ and $I_g$ : $I_2 = 2I_1 - 6I_g$ Substitute this into Eq. 1: $20I_1 + 3I_g - 12(2I_1 - 6I_g) = 0$ $20I_1 + 3I_g - 24I_1 + 72I_g = 0$ $-4I_1 + 75I_g = 0 \implies I_1 = \frac{75}{4}I_g = 18.75I_g$ Now substitute this expression for $I_1$ back into the equation for $I_2$ : $I_2 = 2(18.75I_g) - 6I_g = 37.5I_g - 6I_g = 31.5I_g$ Finally, substitute this expression for $I_2$ into Eq. 3: $13(31.5I_g) + I_g = 2$ $409.5I_g + I_g = 2$ $410.5I_g = 2$ $I_g = \frac{2}{410.5} \approx 0.00487 \text{ A} = 4.87 \text{ mA}$

Final Answer The current through the galvanometer is $4.87 \text{ mA}$ .

Current ElectricityClass 12 Physics Notes

Introduction

Electric Current

Electric Currents in Conductors

Electron Motion in a Conductor

Ohm's Law

Factors Affecting Resistance

Current Density and Vector Form of Ohm's Law

Drift of Electrons and the Origin of Resistivity

Relating Drift Velocity to Current and Resistivity

Given

To Find

Formula

Solution

Solution

Mobility

Limitations of Ohm's Law

Resistivity of Various Materials

Temperature Dependence of Resistivity

Microscopic Explanation

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Electrical Energy, Power

Power Transmission

Cells, emf, Internal Resistance

Cells in Series and in Parallel

Cells in Series

Cells in Parallel

Kirchhoff's Rules

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Wheatstone Bridge

Given

To Find

Formula

Solution