1. Ohm's Law

Ohm's law states that the current passing through a conductor is directly proportional to the potential difference across it, provided temperature remains constant.

$V = IR$

Where:

• $V$ = potential difference (voltage) in volts (V)
• $I$ = current in amperes (A)
• $R$ = resistance in ohms (Ω)

Example 1: A resistor of 4 Ω has a potential difference of 2 V across it. Calculate the current flowing through the resistor.

Solution: $I = \frac{V}{R} = \frac{2V}{4Ω} = 0.5A$

2. Resistance in Series

When resistors are connected in series, the total resistance is the sum of individual resistances:

$R_T = R_1 + R_2 + R_3$

Example 2: Three resistors of 4 Ω, 5 Ω, and 6 Ω are connected in series to a 9 V battery. Find:

• (a) Total resistance
• (b) Current flowing in the circuit

Solution: (a) $R_T = 4 + 5 + 6 = 15Ω$

(b) $I = \frac{V}{R_T} = \frac{9V}{15Ω} = 0.6A$

3. Resistance in Parallel

When resistors are connected in parallel, the reciprocal of total resistance equals the sum of reciprocals:

$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Example 3: Three resistors of 4 Ω, 5 Ω, and 6 Ω are connected in parallel across a 10 V supply. Find the total resistance and total current.

Solution: $\frac{1}{R_T} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{37}{60}$ $R_T = \frac{60}{37} = 1.62Ω$

Total current: $I = \frac{V}{R_T} = \frac{10V}{1.62Ω} = 6.17A$

4. Electric Power

Power is the rate at which electrical energy is transferred:

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

Unit: watt (W)

Example 4: A lamp rated 240 V, 60 W is connected to a 240 V supply. Calculate:

• (a) Current flowing through the lamp
• (b) Resistance of the lamp filament

Solution: (a) $I = \frac{P}{V} = \frac{60W}{240V} = 0.25A$

(b) $R = \frac{V}{I} = \frac{240V}{0.25A} = 960Ω$

5. Electrical Energy and Cost

Energy consumed: $E = P \times t$

Commercial unit: kilowatt-hour (kWh) $1 kWh = 1 kW \times 1 hour = 1000W \times 3600s = 3.6 \times 10^6 J$

Example 5: A television rated 40 W operates 5 hours daily for 30 days. If electricity costs Tsh 229.60 per kWh, calculate the total cost.

Solution: Energy = Power × time = 40 W × 5 h × 30 days = 6000 Wh = 6 kWh

Cost = 6 kWh × Tsh 229.60/kWh = Tsh 1,377.60

1. Mirror Formula (for curved mirrors)

The relationship between object distance ($u$), image distance ($v$), and focal length ($f$):

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Sign convention (New Cartesian):

• Distances measured against incident light are negative
• Distances measured along incident light are positive
• For concave mirrors: $f$ is positive
• For convex mirrors: $f$ is negative

2. Magnification

Magnification relates image size to object size:

$m = \frac{\text{image height}}{\text{object height}} = \frac{h_i}{h_o} = \frac{\text{image distance}}{\text{object distance}} = \frac{v}{u}$

• If $m > 1$: image is enlarged
• If $m < 1$: image is diminished
• If $m$ is negative: image is inverted (real)
• If $m$ is positive: image is upright (virtual)

3. Relationship between Focal Length and Radius of Curvature

$f = \frac{R}{2}$

Example 6: An object 5 cm high is placed 30 cm from a concave mirror with focal length 12 cm. Find:

• (a) Image distance
• (b) Image height and nature

Solution: (a) Using $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$: $\frac{1}{12} = \frac{1}{30} + \frac{1}{v}$ $\frac{1}{v} = \frac{1}{12} - \frac{1}{30} = \frac{5-2}{60} = \frac{3}{60}$ $v = 20 cm$

(b) Magnification: $m = \frac{v}{u} = \frac{20}{30} = \frac{2}{3}$

Image height: $h_i = m \times h_o = \frac{2}{3} \times 5 = 3.33$ cm

Since $m$ is positive but less than 1, the image is virtual, upright, and diminished.

4. Number of Images in Plane Mirrors

When two mirrors are placed at an angle θ: $n = \frac{360°}{θ} - 1$

Example 7: Two plane mirrors are placed at 60° to each other. How many images are formed?

$n = \frac{360°}{60°} - 1 = 6 - 1 = 5$

In Tanzania, these mathematical relations are used daily. For example, when buying electrical appliances like bulbs or irons, you calculate power consumption to estimate electricity costs using the tariff (about Tsh 229-350 per kWh). Similarly, when installing solar panels or setting up home lighting circuits, you apply series and parallel resistance formulas to determine proper wiring and ensure appliances work correctly without overloading the circuit.