Class 11 Basic Electrical Engineering — Exercise Solutions
Key Concepts Quick Recap
| Concept | Formula / Value |
|---|
| Ohm's Law | V = IR; I = V/R; R = V/I |
| Power | P = VI = I²R = V²/R (Watts) |
| Energy | E = Pt (Joules or kWh) |
| Series resistance | R_total = R₁ + R₂ + R₃ |
| Parallel resistance | 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ |
| India AC supply | 230V (rms), 50 Hz, single phase |
| Fuse | Safety device; melts on excess current; in series with circuit |
I. Choose the Best Answer (MCQs)
1. If V = 24V and R = 6Ω, the current I = (a) 2 A (b) 4 A (c) 6 A (d) 144 A
Answer(b) 4 A Ohm's Law: I = V/R = 24/6 = 4 A. Check: P = VI = 24 × 4 = 96 W. Also: P = I²R = 16 × 6 = 96 W. ✓
2. In a series circuit, the total resistance is: (a) Less than the smallest resistance (b) Equal to the sum of all individual resistances (c) Equal to the average of all resistances (d) Always 1 Ω
Answer(b) Equal to the sum of all individual resistances Series: R_total = R₁ + R₂ + ... (resistances add up). Current is same through all. Voltage divides. Parallel: 1/R_total = 1/R₁ + 1/R₂ + ... (total resistance is less than smallest). Voltage is same; current divides.
3. The unit of electrical power is: (a) Joule (b) Coulomb (c) Watt (d) Henry
Answer(c) Watt Power (P) = rate of doing work = VI = I²R = V²/R. Unit: Watt (W) = 1 Joule/second. Energy = Power × Time (Joules or kilowatt-hours, kWh). 1 kWh = 1000 W for 1 hour = 3.6 × 10⁶ J = 1 unit of electricity (for billing).
4. A fuse is connected: (a) In parallel with the load (b) In series with the circuit (c) Between the two live wires (d) After the load
Answer(b) In series with the circuit Fuse: protective device. Thin wire with low melting point; placed in series with circuit. When excess current flows → fuse wire melts → circuit breaks → protects equipment and wiring from damage. Rating should be slightly above normal operating current. MCB (Miniature Circuit Breaker) is modern alternative — auto-resets.
5. Transformer works on the principle of: (a) Mutual induction (b) Self-induction (c) Ohm's law (d) Coulomb's law
Answer(a) Mutual induction Transformer: AC electrical device; primary coil creates changing magnetic flux → induces EMF in secondary coil (mutual induction). Ratio: V₁/V₂ = N₁/N₂ = I₂/I₁. Step-up: N₂ > N₁; V₂ > V₁. Step-down: N₂ < N₁; V₂ < V₁. Only works with AC (DC gives no changing flux → no induction).
6. The resistance of two 6Ω resistors connected in parallel is: (a) 12 Ω (b) 6 Ω (c) 3 Ω (d) 1 Ω
Answer(c) 3 Ω Parallel: 1/R = 1/6 + 1/6 = 2/6 → R = 3 Ω. General rule for two equal resistors in parallel: R_total = R/2. For two unequal: R_total = (R₁ × R₂)/(R₁ + R₂).
7. The standard voltage supply to homes in India is: (a) 110V, 60 Hz (b) 230V, 50 Hz (c) 440V, 50 Hz (d) 220V, 60 Hz
Answer(b) 230V, 50 Hz India: 230V (rms), 50 Hz, single-phase domestic supply. Three-phase supply (for industrial/heavy loads): 415V (line-to-line), 50 Hz. US/Canada: 120V, 60 Hz (single phase). Most of Europe: 230V, 50 Hz (same as India).
II. Short Answer Questions
Q1. Explain the differences between series and parallel circuits with examples.
Answer
Series Circuit:
- Components connected end-to-end in a single path.
- Same current flows through all components.
- Voltage divides across components (V = V₁ + V₂ + V₃).
- R_total = R₁ + R₂ + R₃ (increases).
- If one component fails → circuit breaks → all components stop working.
- Example: old Christmas tree lights (one bulb fails → all go off).
Parallel Circuit:
- Components connected with both ends joined to same two points.
- Voltage is same across all components (same as supply voltage).
- Current divides (I = I₁ + I₂ + I₃).
- R_total = R₁R₂/(R₁+R₂) for two resistors (decreases with each added branch).
- If one component fails → others continue working.
- Example: household wiring (all appliances get 230V; one appliance off doesn't affect others).
Mixed (Series-Parallel): combination of both; solved by simplifying parallel sections first, then total.
Q2. What are the electrical safety precautions to be observed in a workplace?
Answer
Basic electrical safety rules:
1. Never work on live circuits: Always switch off and isolate (LOTO — Lock Out Tag Out) before working on electrical equipment.
2. Use insulated tools: Screwdrivers, pliers, wire cutters should have insulated handles rated for the voltage.
3. Don't overload circuits: Don't connect too many appliances to one socket/extension; check current rating.
4. Use proper fuses/MCBs: Correct rating for each circuit; never replace fuse with a higher-rated one or bypass it.
5. Earthing: All metal-bodied appliances must be properly earthed (third pin of plug). Earthing provides a low-resistance path for fault current → trips fuse/MCB → protects users.
6. Moisture and electricity: Never use electrical equipment with wet hands; keep water away from electrical fittings.
7. Cable management: No frayed or damaged cables; avoid running cables under carpets or through walls without proper conduit.
8. Fire safety: Know location of fire extinguisher (CO₂ type for electrical fires — NOT water).
Exam Checklist
- Ohm's Law: V = IR; I = V/R; R = V/I.
- Power: P = VI = I²R = V²/R (Watts). Energy = P × t (kWh).
- Series: R adds; same I; voltage divides.
- Parallel: R reduces; same V; current divides.
- Fuse: in series; melts on excess current.
- Transformer: mutual induction; V₁/V₂ = N₁/N₂.
- India: 230V, 50 Hz domestic supply.
- Safety: LOTO, earthing, insulated tools, CO₂ fire extinguisher for electrical fires.
