Physics: Class 12 (Book 1) — Chapter 3: Current Electricity

Imagine a small area held normal to the direction of flow of charges. Let $q_+$ be the net positive charge flowing in the forward direction, and $q_-$ be the net negative charge flowing in the forward direction. The net charge flowing across the area in time interval $t$ is:

$$q = q_+ - q_-$$

For a steady flow of charge, the current $I$ is defined as:

$$I = \frac{q}{t}$$

If the rate of flow of charge varies with time, we define the instantaneous current $I(t)$ as:

$$I(t) = \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t} = \frac{dQ}{dt}$$

The SI unit of current is the ampere ( A ). Typical current magnitudes include:

• Domestic appliances: ~ 1 A
• Average lightning: ~ 10,000 A
• Nerves in the human body: ~ 1 &\mu;A

In solid conductors (metals), atoms are tightly bound, and the current is carried by negatively charged electrons moving in the background of fixed positive ions. In electrolytic solutions, both positive and negative ions can move.

In the absence of an electric field, electrons undergo random thermal motion, colliding with fixed ions. The direction of velocity after collision is completely random, resulting in a net average velocity of zero:

$$\frac{1}{N} \sum_{i=1}^{N} \mathbf{v}_i = 0$$

When an electric field $\mathbf{E}$ is applied, electrons experience an electric force and begin to drift, establishing a current.

$$V \propto I \implies V = R I$$

where $R$ is the resistance of the conductor. The SI unit of resistance is the ohm ($\Omega$).

For a conductor of length $l$ and cross-sectional area $A$:

• Resistance is directly proportional to length: $R \propto l$
• Resistance is inversely proportional to cross-sectional area: $R \propto \frac{1}{A}$

Combining these relations gives:

$$R = \rho \frac{l}{A}$$

where $\rho$ is the resistivity of the material. Resistivity depends on the material type and temperature, but not on its dimensions.

Define current density $j$ as the current per unit area normal to the flow:

$$j = \frac{I}{A}$$

The SI unit of current density is A/m² . If $E$ is the uniform electric field in a conductor of length $l$, then $V = El$. Substituting these into Ohm's law:

$$El = I R = I \left(\rho \frac{l}{A}\right) = \left(\frac{I}{A}\right) \rho l = j \rho l$$

$$E = j \rho \implies \mathbf{E} = \mathbf{j} \rho$$

Alternatively, in terms of conductivity $\sigma = 1/\rho$:

$$\mathbf{j} = \sigma \mathbf{E}$$

Let $\tau$ be the relaxation time (the average time between successive collisions). The velocity of the $i$-th electron at time $t$ is:

$$\mathbf{V}_i = \mathbf{v}_i - \frac{e\mathbf{E}}{m} t_i$$

Averaging over all $N$ electrons, the average initial thermal velocity $\langle \mathbf{v}_i \rangle = 0$, and the average time elapsed since the last collision is $\tau$:

$$\mathbf{v}_d = -\frac{e\mathbf{E}}{m} \tau$$

Consider a conductor of cross-sectional area $A$. Let $n$ be the number density of free electrons. In a time interval $\Delta t$, the electrons drift a distance $|v_d| \Delta t$. The total charge crossing the area $A$ in time $\Delta t$ is:

$$\Delta Q = n e A |v_d| \Delta t$$

The current $I$ is therefore:

$$I = n e A v_d$$

Substituting the expression for $v_d$ into the current equation:

$$I = n e A \left(\frac{e E \tau}{m}\right) = \frac{n e^2 A \tau}{m} E$$

Since $j = I/A$ and $E = V/l$:

$$j = \left(\frac{n e^2 \tau}{m}\right) E$$

Comparing this with $\mathbf{j} = \sigma \mathbf{E}$, we get the microscopic expressions for conductivity ($\sigma$) and resistivity ($\rho$):

$$\sigma = \frac{n e^2 \tau}{m}$$

$$\rho = \frac{m}{n e^2 \tau}$$

Mobility ($\mu$) is defined as the magnitude of drift velocity per unit electric field:

$$\mu = \frac{|v_d|}{E} = \frac{e \tau}{m}$$

The SI unit of mobility is m² V¹ s¹ .

• Non-linear $V-I$ relation: At high currents, heating effects cause resistance to increase, making the $V-I$ curve non-linear.
• Directional dependence: In devices like semiconductor diodes, the current magnitude depends on the sign of the applied voltage.
• Non-unique relation: In materials like Gallium Arsenide (GaAs), there can be more than one value of voltage for the same current, exhibiting negative resistance regions.

Over a limited temperature range, the resistivity of a metallic conductor is given by:

$$\rho_T = \rho_0 [1 + \alpha(T - T_0)]$$

where $\alpha$ is the temperature coefficient of resistivity , measured in °C¹ or K¹ .

• Metals: $\alpha$ is positive. Resistivity increases with temperature because thermal vibrations increase, decreasing the relaxation time $\tau$.
• Semiconductors & Insulators: Resistivity decreases exponentially with temperature because the carrier concentration $n$ increases rapidly, overcoming the slight decrease in $\tau$.
• Alloys (Nichrome, Manganin, Constantan): Exhibit very weak temperature dependence, making them ideal for standard wire-bound resistors.

The energy dissipated as heat in a conductor of resistance $R$ in time interval $\Delta t$ is:

$$\Delta W = I V \Delta t$$

The power $P$ dissipated is:

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

In power transmission, to minimize power loss $P_c$ in transmission lines of resistance $R_c$, electricity is transmitted at very high voltage $V$:

$$P_c = \frac{P^2 R_c}{V^2}$$

An electrolytic cell maintains a steady current by converting chemical energy. It consists of two electrodes immersed in an electrolyte.

• Electromotive Force (emf, $\varepsilon$): The potential difference between the positive and negative electrodes in an open circuit (when no current flows).
• Internal Resistance ($r$): The resistance offered by the electrolyte to the flow of current inside the cell.
• Terminal Voltage ($V$): The potential difference across the terminals of the cell when current $I$ is flowing:$$V = \varepsilon - I r$$
• Maximum Current: The maximum current that can be drawn from a cell (when external resistance $R = 0$) is:$$I_{max} = \frac{\varepsilon}{r}$$

For two cells of emfs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ connected in series:

$$V_{AC} = V_A - V_C = (\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2)$$

The equivalent emf and internal resistance are:

$$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$$

$$r_{eq} = r_1 + r_2$$

If one cell is connected with reverse polarity:

$$\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2 \quad (\text{for } \varepsilon_1 > \varepsilon_2)$$

For two cells connected in parallel:

$$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$$

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

In a simpler form:

$$\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}$$

$$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$$

1. Junction Rule (First Rule): At any junction, the sum of currents entering the junction must equal the sum of currents leaving it. This is based on the conservation of charge .$$\sum I_{in} = \sum I_{out}$$
2. Loop Rule (Second Rule): The algebraic sum of changes in potential around any closed loop involving resistors and cells must be zero. This is based on the conservation of energy .$$\sum \Delta V = 0$$

The Wheatstone bridge is an arrangement of four resistors $R_1, R_2, R_3, R_4$ in a loop with a galvanometer $G$ connected across one diagonal and a source across the other.

Under the balanced condition, no current flows through the galvanometer ($I_g = 0$). Applying Kirchhoff's rules yields the balance condition:

$$\frac{R_2}{R_1} = \frac{R_4}{R_3}$$

This principle is used in practical devices like the meter bridge to measure unknown resistances.