Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant:

$I = \frac{V}{R} \quad \text{or} \quad V = IR$

where $R$ is the resistance measured in ohms ($\Omega$).

Resistivity

The resistance of a uniform conductor depends on its dimensions and the material's property called resistivity ($\rho$):

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

where:

• $l$ = length of the conductor
• $A$ = cross-sectional area
• $\rho$ = resistivity (Ω·m)

Conductivity ($\sigma$) is the reciprocal of resistivity: $\sigma = \frac{1}{\rho}$

Temperature Dependence

Resistance changes with temperature according to: $R_\theta = R_o[1 + \alpha(\theta - \theta_o)]$

where $\alpha$ is the temperature coefficient of resistance.

Example 2.1: A copper wire has a resistance of 10 Ω at 20°C. Find its resistance at 60°C, given $\alpha = 0.004$/°C.

Solution: $R_{60} = 10[1 + 0.004(60 - 20)] = 10[1 + 0.16] = 11.6\ \Omega$

Electromotive Force and Internal Resistance

A source of emf (like a battery) provides energy to move charges through a circuit. When current flows, some voltage is lost across the internal resistance ($r$) of the source:

$V = E - Ir$

where:

• $E$ = emf of the source
• $V$ = terminal voltage
• $r$ = internal resistance

Series and Parallel Combinations

Series combination: $R_{eq} = R_1 + R_2 + R_3 + \cdots$

Parallel combination: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$

Kirchhoff's Laws

Kirchhoff's Current Law (KCL): At any junction, the algebraic sum of currents is zero: $\sum I = 0$

Kirchhoff's Voltage Law (KVL): In any closed loop, the algebraic sum of emfs equals the algebraic sum of potential drops: $\sum E = \sum IR$

AC Voltage and Current

In AC circuits, voltage and current vary sinusoidally with time: $E = E_o \sin\omega t = E_o \sin 2\pi ft$ $I = I_o \sin\omega t$

where:

• $E_o$, $I_o$ = peak values
• $f$ = frequency (Hz)
• $\omega$ = angular frequency = $2\pi f$

Root Mean Square (rms) Values

Since AC values change continuously, we use rms values for practical measurement:

$I_{rms} = \frac{I_o}{\sqrt{2}} = 0.707I_o$ $E_{rms} = \frac{E_o}{\sqrt{2}} = 0.707E_o$

The rms value represents the equivalent DC value that would produce the same heating effect.

AC Circuit Elements

Resistor only: Voltage and current are in phase.

Inductor only: Current lags voltage by 90° (π/2). Inductive reactance: $X_L = \omega L = 2\pi fL$

Capacitor only: Current leads voltage by 90° (π/2). Capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}$

Series RLC Circuits

The impedance is: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

The phase angle: $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$

Resonance occurs when $X_L = X_C$, giving: $f_o = \frac{1}{2\pi\sqrt{LC}}$

At resonance, impedance is minimum (equal to $R$) and current is maximum.

Power in AC Circuits

Average power: $P_{av} = I_{rms} E_{rms} \cos\phi$

The power factor is $\cos\phi$, where $\phi$ is the phase angle between voltage and current.

Example 4.1: A 60 Ω resistor is connected to a 240 V, 50 Hz AC supply. Find: (a) the rms current (b) the peak voltage

Solution: (a) $I_{rms} = \frac{V_{rms}}{R} = \frac{240}{60} = 4$ A

(b) $E_o = E_{rms} \times \sqrt{2} = 240 \times 1.414 = 339.4$ V

Wheatstone Bridge

A balanced Wheatstone bridge has: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$

For finding unknown resistance $R_x$: $R_x = \frac{R_3}{R_4} \times R_2$

Potentiometer

The potentiometer measures emf by balancing it against a known potential drop along a uniform wire. At balance: $E_x \propto l_x$

For comparing emfs: $\frac{E_x}{E_s} = \frac{l_x}{l_s}$

Internal resistance of a cell: $r = \left(\frac{l - l_1}{l_1}\right)R$

In Tanzania, understanding current electricity is essential for safely using electrical appliances in homes and businesses. For instance, when connecting a 2,000 W rice cooker (rated 240 V) in Dar es Salaam, knowing how to calculate current ($I = P/V = 2000/240 \approx 8.3$ A) helps in selecting the appropriate fuse or circuit breaker to prevent overloading and fire hazards. Similarly, solar home systems increasingly used in rural areas rely on understanding DC circuits and battery charging to ensure reliable off-grid electricity supply.