Electricity · Lesson 9 of 9
Chapter Summary and Practice
“The chapter review reconnects every electrical idea without causing a short circuit in your memory.”
• Connect current, potential difference, resistance, heating, power and energy. • Recall the important electrical definitions and circuit rules. • Select formulas from the quantities given in a problem. • Reduce series, parallel and mixed resistor arrangements. • Check calculations using units, limiting cases and physical meaning. • Apply chapter ideas to appliances, heating and electrical energy use.
Electricity links the movement of charge with the transfer of energy. A source maintains potential difference, a complete circuit permits current, and resistance controls how much current flows. The same current can produce heating, light or mechanical work, while power describes how quickly energy is transferred. These ideas become one connected system when circuit rules and units are used carefully.
Current, Charge and Circuits
Electric current is charge passing per unit time, I = Q/t. In metal wires, electrons drift from the negative terminal toward the positive terminal, while conventional current is taken in the opposite direction. A closed conducting path is required for steady current. An ammeter is connected in series because it must carry the current being measured.
Potential difference is energy transferred per unit charge, V = W/Q. A cell or battery maintains this difference by converting chemical energy. A voltmeter is connected in parallel across two points. Current can be zero even when a potential difference exists if the path is open.
Circuit Symbols and Measurement
Conventional symbols show electrical connections rather than realistic appearance. Long and short lines identify cell polarity, an open key shows a broken path, and a junction dot shows joined wires. The ammeter belongs in the current path; the voltmeter bridges the component whose potential difference is required.
Ohm’s Law and Resistance
For an ohmic conductor at constant temperature, V = IR. Resistance is V/I and is measured in ohms. On a graph of V against I, resistance is the slope. A rheostat varies resistance to control current. The constant-temperature condition matters because heating can change resistance.
For a uniform wire, R = ρl/A. Resistance belongs to the particular wire; resistivity characterises its material at a specified temperature. Longer wires have greater resistance and thicker wires smaller resistance. Because A = πd²/4, resistance varies inversely with the square of diameter.
Series and Parallel Circuits
Series resistors share the same current and divide the total potential difference. Their equivalent resistance is the sum Rₛ = R₁ + R₂ + ···. Parallel resistors share the same potential difference and divide current; their equivalent follows 1/Rₚ = 1/R₁ + 1/R₂ + ···. A parallel equivalent is always below the smallest branch resistance.
Mixed circuits should be reduced from clearly identifiable subgroups. Replace one series or parallel group by its equivalent, redraw mentally, and continue. Appliances are connected in parallel so each receives the full supply voltage and can operate independently.
Heating, Power and Energy
Electrical energy transferred in time t is VIt. In a purely resistive component it appears as heat: H = VIt = I²Rt = V²t/R. Joule heating explains irons, heaters, incandescent filaments and fuse operation. A fuse is placed in series and melts when excessive current produces rapid heating.
Power is the rate of energy transfer: P = VI = I²R = V²/R. Energy is E = Pt. Watts multiplied by seconds give joules, while kilowatts multiplied by hours give kilowatt-hours. One kilowatt-hour equals 3.6 × 10⁶ J. Energy cost is calculated from kilowatt-hours, not from power alone.
Important Definitions
| Term | Meaning |
|---|---|
| Electric current | Rate of flow of electric charge |
| Electric circuit | Continuous conducting path for current |
| Potential difference | Work done or energy transferred per unit charge |
| Resistance | Opposition offered by a particular conductor to current |
| Resistivity | Material property represented by ρ = RA/l |
| Ohm’s law | V is proportional to I at constant temperature |
| Series combination | Components sharing one unbranched path |
| Parallel combination | Components connected between the same two junctions |
| Heating effect | Conversion of electrical energy into heat in a conductor |
| Electric power | Rate of electrical energy transfer |
| Watt | One joule per second or one volt ampere |
| Kilowatt-hour | Energy used by one kilowatt operating for one hour |
Formula Review
A Unified Problem-Solving Method
- Translate the situation into a circuit or a list of electrical quantities.
- Convert mA to A, minutes to seconds when needed, mm to m, and W to kW for billing.
- Identify series and parallel groups from junctions, then find equivalent resistance.
- Use total voltage and equivalent resistance to find total current.
- Return to individual branches to find branch currents, voltage drops, power or heat.
- Check series voltage sums, parallel current sums, resistance limits and units.
- Interpret the result physically instead of reporting an unexplained number.
Common Conceptual Mistakes
| Mistake | Correction |
|---|---|
| Treating current as energy | Current is charge per time; energy transferred is measured in joules or kilowatt-hours |
| Connecting meters the same way | Ammeter is series; voltmeter is parallel |
| Ignoring temperature in Ohm’s law | V/I is constant only under unchanged physical conditions for an ohmic conductor |
| Using diameter as area | For a round wire use A = πd²/4 |
| Adding parallel resistances directly | Add reciprocals and invert |
| Expecting parallel resistance to be large | Extra paths make it smaller than the smallest branch |
| Calling kWh a power unit | kWh is energy; kW is power |
| Assuming electrons are consumed | Charge circulates while energy is transferred |
Mixed Worked Examples
Problem
A 6 Ω resistor is in series with a parallel pair of 12 Ω and 4 Ω across 18 V. Find total current and current in each parallel branch.
- 1.Parallel group: Rₚ = (12 × 4)/(12 + 4) = 48/16 = 3 Ω.
- 2.Total resistance: R = 6 + 3 = 9 Ω.
- 3.Total current: I = 18/9 = 2 A.
- 4.Voltage across series 6 Ω resistor: 2 × 6 = 12 V, leaving 18 − 12 = 6 V across the parallel group.
- 5.Branch currents: I₁₂ = 6/12 = 0.5 A and I₄ = 6/4 = 1.5 A.
- 6.Check: 0.5 + 1.5 = 2 A, equal to total current.
Problem
A 44 Ω heater operates at 220 V for 2 h. Find current, power and energy in kWh.
- 1.Current: I = V/R = 220/44 = 5 A.
- 2.Power: P = VI = 220 × 5 = 1100 W = 1.1 kW.
- 3.Energy: E = Pt = 1.1 kW × 2 h = 2.2 kWh.
- 4.In joules, E = 2.2 × 3.6 × 10⁶ = 7.92 × 10⁶ J.
- 5.The units confirm that power is a rate and energy accumulates over time.
Problem
Two identical 6 Ω resistors are connected first in series and then in parallel across the same voltage for the same time. Compare total heat.
- 1.Series equivalent: Rₛ = 6 + 6 = 12 Ω.
- 2.Parallel equivalent: Rₚ = (6 × 6)/(6 + 6) = 3 Ω.
- 3.At fixed voltage and time, total heat H = V²t/R.
- 4.Hseries/Hparallel = Rparallel/Rseries = 3/12 = 1/4.
- 5.Thus series:parallel heat = 1:4. Parallel draws greater current because its equivalent resistance is smaller.
Quiz
Which pair correctly matches instrument and connection?
A wire is cut to half its length without changing area. What happens to its resistance?
Which result can be an equivalent resistance of 6 Ω and 3 Ω in parallel?
Which quantity is measured in kilowatt-hours?
At fixed voltage, which resistor consumes more power?
Practice Problems
- A current of 1.5 A flows for 8 min. Find charge. Solution: t = 480 s. Q = It = 1.5 × 480 = 720 C.
- A 24 V source transfers 960 J. Find charge. Solution: Q = W/V = 960/24 = 40 C.
- A wire of resistance 8 Ω has its length doubled and area halved. Find new resistance. Solution: R₂/R₁ = (2)(2) = 4, so R₂ = 32 Ω.
- Find the equivalent resistance of 2 Ω and 4 Ω in series, connected in parallel with 3 Ω. Solution: Series branch = 6 Ω. Parallel with 3 Ω gives R = (6 × 3)/(6 + 3) = 2 Ω.
- The network in the previous problem is across 12 V. Find total current. Solution: I = V/R = 12/2 = 6 A.
- A 20 Ω resistor carries 3 A for 2 min. Find heat. Solution: t = 120 s. H = I²Rt = 9 × 20 × 120 = 21,600 J.
- A 220 V appliance draws 4 A. Find power and energy used in 3 h. Solution: P = VI = 880 W = 0.88 kW. E = 0.88 × 3 = 2.64 kWh.
- How many 176 Ω resistors in parallel are needed to carry 5 A from 220 V? Solution: Required equivalent R = V/I = 220/5 = 44 Ω. For n identical parallel resistors, Rₚ = 176/n, so n = 176/44 = 4.
- A 100 Ω lamp, 50 Ω toaster and 500 Ω filter are parallel across 220 V. Find total current. Solution: Branch currents are 2.2 A, 4.4 A and 0.44 A. Total = 7.04 A.
- A 250 W television runs 4 h and a 1 kW heater runs 45 min. Which uses more energy? Solution: Television = 0.250 × 4 = 1.0 kWh. Heater = 1 × 0.75 = 0.75 kWh. The television uses 0.25 kWh more.
Key Takeaways
• Current describes charge flow; potential difference describes energy per charge. • Ohm’s law links voltage, current and resistance at constant temperature. • Wire resistance depends on resistivity, length and cross-sectional area. • Series circuits share current; parallel circuits share voltage. • Joule heating depends on current squared, resistance and time. • Power is energy transfer per second; energy equals power multiplied by time. • One kilowatt-hour equals 3.6 × 10⁶ joules. • Reliable solutions use correct connections, units, resistance limits and physical checks.
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Electric Power
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