Lesson 2: Electric Circuits

2.1 How is the current in a conductor determined? Ohm's law

• In a closed circuit a current flows through a conductor when a potential difference is applied across it, and the conductor's resistance opposes the flow.
• How are the three quantities related?

Activity: verifying Ohm's law.

• Apparatus: battery, fixed resistor, rheostat, switch, ammeter, voltmeter, wires.
• Connect the ammeter in series with the resistor and rheostat, and the voltmeter across the fixed resistor.

Steps:

1. Set the rheostat at its maximum value so that the current is least (this protects the components).
2. Close the circuit and record the voltmeter reading V and the ammeter reading I.
3. Reduce the rheostat resistance; record the new V and I.
4. Repeat several times and tabulate the readings.
5. Plot a graph of V (horizontal axis) against I (vertical axis).

Variables: the independent variable is the potential difference across the resistor (changed by the rheostat); the dependent variable is the current through the resistor; the controlled (constant) factors are the resistor itself (its value), its temperature and the materials.

Result: the graph is a straight line passing through the origin, which means that I is directly proportional to V.

Ohm's law: the current intensity through a conductor (fixed resistor) is directly proportional to the potential difference across its ends, at constant temperature.

I = V / R, or V = I R, or R = V / I

The resistance can be found from the graph: R = ΔV / ΔI, which is the reciprocal of the slope of an I–V graph (where I is plotted on the vertical axis, slope = I/V = 1/R, so R = 1/slope).

Scientist profile. Georg Simon Ohm (1789–1854), a German physicist who found the law named after him and the relationship between current, voltage and resistance; the unit of resistance is named in his honour.

Worked example 1

Find the current in a metal wire of resistance 2 Ω when the potential difference across it is 6 V. I = V/R = 6 / 2 = 3 A.

Worked example 2: from a graph

An I–V graph for a fixed resistor is a straight line through the origin passing through the point (V = 2.0 V, I = 0.40 A). Find R. R = V/I = 2.0 / 0.40 = 5 Ω. Check with another point: at 1.0 V the current should be 0.20 A.

Worked example 3

What potential difference is needed to drive 0.25 A through a 48 Ω resistor? V = I R = 0.25 × 48 = 12 V.

What happens if the potential difference doubles? The resistance of the conductor does not change (it depends on length, area, material and temperature); the current doubles.

2.2 Connecting identical resistors

Several resistors may be connected in two ways.

Point of comparison	In series	In parallel
Path	One single path	Branching paths
Current	Same current through every resistor	Current divides between the branches (equally for identical resistors); total current from the source = sum of branch currents
Potential difference	Divided among the resistors (equally for identical resistors); sum of the parts = source emf	The same across every resistor (equal to the emf)
Equivalent resistance, R′, for n identical resistors each R	R′ = n R	R′ = R / n
Compared with one resistor	Larger	Smaller
Purpose	To get a large resistance	To get a small resistance

The equivalent resistance (R′) is the single resistance that has the same effect on the circuit as the whole group of resistors.

Life application.

• Household appliances are connected in parallel to the mains, so that each can be switched on or off independently, and each gets the full mains voltage.
• If one lamp burns out, the others keep working, since each has its own path to the source.

Worked example 4: series

Three identical resistors are connected in series to a 12 V source. The current is 2 A. Find the resistance of each. Total (equivalent) resistance R′ = V / I = 12 / 2 = 6 Ω. R′ = 3R, so R = 6/3 = 2 Ω. Check: each resistor has 4 V across it (2 A × 2 Ω), and 3 × 4 = 12 V.

Worked example 5: parallel

Three identical 6 Ω resistors are connected in parallel to a 12 V battery. Find the equivalent resistance, the total current, and the current in each resistor. R′ = R/n = 6/3 = 2 Ω. Total current I = 12/2 = 6 A. Each branch: 12/6 = 2 A, and 3 × 2 = 6 A.

Worked example 6: two ways with the same resistors

Two identical 10 Ω resistors are connected to a 10 V battery, first in series, then in parallel. In series R′ = 20 Ω, I = 0.5 A. In parallel R′ = 5 Ω, I = 2 A. The parallel arrangement gives a current four times as large. This is why, if you want the maximum reading on an ammeter with two identical resistors and two identical cells, you connect the cells in series (to double the emf) and the resistors in parallel (to halve the resistance).

• Pacemaker (integration with medicine). A pacemaker is a small electronic circuit implanted under the skin: a battery, a microchip, and wires to the heart muscle.
• If the heartbeat becomes irregular, the circuit sends tiny electric pulses to make the heart contract regularly.

2.3 Electric power and energy

Energy-efficiency label. Appliances such as refrigerators carry a label that shows the energy rating and the monthly energy consumption, so that buyers can choose appliances that consume less (the highest rating, shown by the green arrow, A, means the highest efficiency and lowest consumption).

Electric power (P) is the rate at which an appliance consumes electric energy, i.e. the energy it consumes per second.

P = E / t, so E = P × t

Unit of power: watt (W) = 1 joule per second. "A 9 W LED lamp" uses 9 joules of electrical energy every second.

The joule is a very small unit of energy, so in household billing the larger unit kilowatt-hour (kW·h) is used: 1 kW·h = 1000 W × 3600 s = 3.6 × 10⁶ J.

Cost of electric energy. Follow four steps:

1. Power in kW = power in W ÷ 1000.
2. Find the number of operating hours.
3. Energy in kW·h = power in kW × hours.
4. Cost = energy in kW·h × price per kW·h.

(The price per kW·h used in the book's example is 1.55 Egyptian pounds. For your own calculations use the price given in the question.)

Worked example 7: comparing two lamps

A CFL of 20 W and an LED of 12 W give about the same light. Compare the monthly (30-day) costs if each is used 12 h per day at 1.55 pounds per kW·h. Hours = 30 × 12 = 360 h. CFL: 0.020 kW × 360 = 7.2 kW·h; cost = 7.2 × 1.55 = 11.16 pounds. LED: 0.012 × 360 = 4.32 kW·h; cost = 4.32 × 1.55 = 6.70 pounds. Saving = about 4.46 pounds per month per lamp. The LED is more efficient.

Worked example 8: a kettle

A 1500 W kettle is used 0.5 h per day. What is the weekly energy and cost at 1.55 pounds per kW·h? P = 1.5 kW. Hours per week = 0.5 × 7 = 3.5 h. Energy = 1.5 × 3.5 = 5.25 kW·h. Cost = 5.25 × 1.55 = 8.1375, about 8.14 pounds.

Worked example 9: equal energy

Appliance X has power 2000 W, appliance Y has 1000 W. X runs for 3 hours. How long must Y run to use the same energy? E = 2 kW × 3 h = 6 kW·h; time for Y = 6 / 1 = 6 hours. The lower the power, the longer it takes to use the same energy.

• Typical powers (for reference): electric heater 1–2 kW; iron 1–2.5 kW; air conditioner 1.5–2.5 kW; laptop about 50 W; fan 50–100 W; television about 100 W.
• Heating appliances and air conditioners consume the most energy.
• Increasing the operating hours (cause) increases the energy and the cost (effect).

2.4 Electrical safety

• Avoid overloading: do not plug several high-power appliances into one socket; the large current can overheat the wires and cause a fire.
• Insulation: wires must be properly insulated so that you cannot get a shock.
• Dryness: make sure hands, wires and components are dry; tap water conducts electricity.
• Disconnect the source before changing anything in a circuit.
• Never use household mains in school experiments; use batteries.

2.5 Summary of Lesson 2

• Ohm's law: I ∝ V at constant temperature; R = V/I; the graph is a straight line through the origin.
• Series: same current, voltage divides, R′ = nR. Parallel: same voltage, current divides, R′ = R/n.
• P = E/t; E (kW·h) = P (kW) × t (h); cost = E × price.