Current Electricity and Ohm's Law

Basic quantities

• Electric current (I): rate of flow of charge, I = Q/t, in amperes (A). One ampere is one coulomb per second.
• Potential difference (V): energy transferred per unit charge, V = W/Q, in volts (V). Measured with a voltmeter connected in parallel.
• Resistance (R): opposition to current, in ohms. Measured as R = V/I.
• An ammeter is connected in series.

In Kenya mains electricity is about 240 V at 50 Hz, supplied by KPLC from sources such as geothermal and hydro stations.

Ohm's law

At constant temperature the current through a conductor is directly proportional to the potential difference across it:

V = IR

A graph of V against I for an ohmic conductor is a straight line through the origin; its gradient is the resistance. A filament lamp is non-ohmic because its resistance rises as it heats.

Worked example 1: A resistor carries 0.5 A when connected to 6 V. R = 6 / 0.5 = 12 ohms.

Factors affecting resistance of a wire

Resistance increases with length, decreases with cross-sectional area, depends on the material, and generally rises with temperature. R = (resistivity x length) / area.

Series circuits

• The same current flows through every component.
• Voltages add: V = V1 + V2 + ...
• Total resistance R = R1 + R2 + ...

Parallel circuits

• Voltage is the same across each branch.
• Currents add: I = I1 + I2 + ...
• 1/R = 1/R1 + 1/R2 + ...; for two resistors, R = R1R2 / (R1 + R2).

Worked example 2: A 6 ohm and 3 ohm resistor are in parallel across 12 V.

• R = (6 x 3)/9 = 2 ohms
• Current in 6 ohm = 12/6 = 2 A; in 3 ohm = 12/3 = 4 A
• Total current = 6 A (check: 12/2 = 6 A)

Electromotive force and internal resistance

The e.m.f. (E) of a cell is the total energy supplied per coulomb. E = V + Ir, where r is the internal resistance, so the terminal voltage drops when current is drawn.

Electrical power and energy

• P = VI = I^2R = V^2/R (watts)
• Energy W = Pt (joules), or kWh = kW x hours

Worked example 3: A 2 kW kettle is used 15 minutes daily for 30 days. Energy = 2 x 0.25 x 30 = 15 kWh. At KES 25/kWh, cost = KES 375.

Fuses and safety

A fuse protects against overcurrent. Choose a fuse rating just above normal current: I = P/V. For a 460 W appliance at 230 V, I = 2 A, so use a 3 A fuse. Earth wires, circuit breakers and insulated cables prevent shock and fire.

Try it

• Find the current in a 20 ohm resistor at 10 V.
• Calculate the total resistance of 5, 10 and 15 ohms in series.
• Two 12 ohm resistors are in parallel; find the total resistance.
• Calculate the monthly cost of a 100 W bulb on for 5 hours a day at KES 25/kWh.