Chapter 11: Electricity
Key Formulas:
• Ohm's Law: V = IR
• Resistance: R = ρl/A
• Series: Rₑ = R₁ + R₂ + R₃
• Parallel: 1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃
• Power: P = VI = I²R = V²/R
• Energy: E = Pt = VIt
Q1. What does an electric circuit mean?
Solution
An electric circuit is a continuous and closed path through which electric current flows. It consists of a source of electricity (battery), wires, and components like resistors, bulbs, and switches. When the circuit is complete (closed), current flows.
Q2. Define the unit of electric current.
Solution
The SI unit of electric current is the Ampere (A). One ampere is defined as the flow of one coulomb of charge per second through any cross-section of a conductor.
Q3. Name the device that helps to maintain a potential difference across a conductor.
Solution
A cell or battery helps to maintain a potential difference across a conductor. A battery is a combination of two or more cells connected together.
Q4. State the factors on which the resistance of a conductor depends.
Solution
Resistance depends on:
• Length (l): R is directly proportional to length
• Cross-sectional area (A): R is inversely proportional to area
• Material: Different materials have different resistivity (ρ)
• Temperature: Resistance generally increases with temperature
Q5. A wire of resistance R is doubled upon itself. What is the new resistance?
Solution
When doubled: l' = l/2, A' = 2A
R' = ρl'/A' = ρ(l/2)/(2A) = ρl/(4A) = R/4
New resistance = R/4
Q6. An electric bulb is connected to a 220V generator. If the current is 0.50A, what is the power of the bulb?
Solution
P = V × I
P = 220 × 0.50
P = 110 W
The power of the bulb is 110 Watts.
Q7. Two lamps, one rated 100 W at 220V, and the other 60 W at 220V, are connected in parallel to electric mains. What is the current drawn from the line if the supply voltage is 220V?
Solution
For 100 W lamp: I₁ = P/V = 100/220 = 0.455 A
For 60 W lamp: I₂ = P/V = 60/220 = 0.273 A
Total current = I₁ + I₂ = 0.455 + 0.273 = 0.727 A
The total current drawn is 0.727 A.
Q8. Calculate the resistance of a metal wire of length 2 m and area of cross-section 1.55 × 10⁻⁶ m². The resistivity of the material is 2.8 × 10⁻⁸ Ωm.
Solution
R = ρl/A
R = (2.8 × 10⁻⁸) × 2 / (1.55 × 10⁻⁶)
R = 5.6 × 10⁻⁸ / 1.55 × 10⁻⁶
R = 3.61 × 10⁻⁸ Ω
R = 0.036 Ω
The resistance is 0.036 Ω.
Q9. An electric heater of resistance 8 Ω draws 15 A current from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution
P = I²R
P = (15)² × 8
P = 225 × 8 = 1800 W
P = 1800 J/s
The rate of heat development is 1800 J/s (1800 Watts).
Q10. How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?
Solution
Total resistance needed: Rₑ = V/I = 220/5 = 44 Ω
For n identical resistors in parallel:
Rₑ = R/n
44 = 176/n
n = 176/44 = 4
4 resistors of 176 Ω are needed in parallel.
Q11. Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω
Solution
(i) For 9 Ω:
Two 6 Ω in series = 12 Ω... No.
Better: Two in parallel (3 Ω) + one in series = 3 + 6 = 9 Ω ✓
(R₁ ∩ R₂) + R₃ = (6×6)/(6+6) + 6 = 3 + 6 = 9 Ω
(ii) For 4 Ω:
Three in parallel: 1/Rₑ = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Rₑ = 2 Ω... No.
Two in series (12 Ω) in parallel with one 6 Ω: (12×6)/(12+6) = 72/18 = 4 Ω ✓
(i) Two in parallel + one in series = 9 Ω
(ii) Two in series, all three in parallel = 4 Ω
Q12. What is the commercial unit of electrical energy? Convert it into joules.
Solution
1 kilowatt-hour (kWh) = 1000 W × 3600 s = 3.6 × 10⁶ J
The commercial unit is kilowatt-hour (kWh). 1 kWh = 3.6 × 10⁶ Joules.