Electrical Power and Household Circuits — ICSE Class 10 Physics Important Questions

Answer: (c) kilowatt-hour.

Electrical energy is sold in kilowatt-hour (kWh), where $1\text{ kWh}=3.6\times10^{6}\text{ J}$1 kWh=3.6×10^6 J. The watt is the unit of power and the joule is the SI unit of energy.

Answer: (c) live wire.

Both the fuse and the switch are placed in the live wire so that when they operate, the appliance is disconnected from the high-potential live line, making it safe to touch.

Answer: (b) $100\,\text{W}$100\,W lamp.

At the same voltage, $I=\dfrac{P}{V}$I=P/V, so the higher-power lamp draws more current: $I_{100}=\dfrac{100}{220}\approx0.45\text{ A}$I_100=100/2200.45 A versus $I_{60}=\dfrac{60}{220}\approx0.27\text{ A}$I_60=60/2200.27 A. Hence the $100\text{ W}$100 W lamp glows brighter.

Answer: (c) $250\,\text{W}$250\,W.

Resistance is fixed: $R=\dfrac{V^2}{P}=\dfrac{220^2}{1000}=48.4\,\Omega$R=V^2/P=220^2/1000=48.4\,. At $110\text{ V}$110 V, $P'=\dfrac{V'^2}{R}=\dfrac{110^2}{48.4}=250\text{ W}$P'=V'^2/R=110^2/48.4=250 W. Since $P\propto V^2$P V^2, halving $V$V quarters the power.

Answer: (a) The metal body is earthed, and earthing indeed gives leakage/fault current a safe low-resistance path to the ground, so the fuse blows and the user is protected. R correctly explains A.

The kilowatt-hour (kWh) is the electrical energy consumed by an appliance of power $1\text{ kW}$1 kW used for $1\text{ hour}$1 hour. It is the commercial unit of electrical energy.

$1\text{ kWh}=1000\text{ W}\times3600\text{ s}=3.6\times10^{6}\text{ J}.$1 kWh=1000 W×3600 s=3.6×10^6 J.

Fuse:

1. It is a safety device that melts and breaks the circuit automatically when the current exceeds a safe value.
2. Once blown, it must be replaced (or the MCB reset) before use.

Switch:

1. It is a manually operated device used to make or break the circuit at will.
2. It is not destroyed in operation and can be reused repeatedly.

Both are connected in the live wire.

Given: $P=1500\text{ W}$P=1500 W, $V=220\text{ V}$V=220 V.

(i) Current:
$I=\dfrac{P}{V}=\dfrac{1500}{220}=6.82\text{ A}.$I=P/V=1500/220=6.82 A.

(ii) Resistance:
$R=\dfrac{V^2}{P}=\dfrac{220^2}{1500}=\dfrac{48400}{1500}=32.3\,\Omega.$R=V^2/P=220^2/1500=48400/1500=32.3\,.

(iii) Energy in 30 days:
Daily energy $=\dfrac{1500}{1000}\times2=3\text{ kWh}$=1500/1000×2=3 kWh.
$E=3\times30=90\text{ kWh}.$E=3×30=90 kWh.

A three-core cable has three insulated wires:

1. Live wire — brown (older code: red). It carries current at high potential from the supply to the appliance. The fuse and switch are placed in this wire.

2. Neutral wire — blue (older code: black). It is at (nearly) earth potential and completes the circuit, carrying current back to the supply.

3. Earth wire — green with yellow stripes (older code: green). It connects the metal body of the appliance to the ground and provides a safe path for leakage current, protecting the user from shock.

Lamps: energy $=\dfrac{5\times60}{1000}\times5=\dfrac{300}{1000}\times5=1.5\text{ kWh}$=5×60/1000×5=300/1000×5=1.5 kWh.

Heater: energy $=\dfrac{1000}{1000}\times2=2\text{ kWh}$=1000/1000×2=2 kWh.

Refrigerator: energy $=\dfrac{200}{1000}\times10=2\text{ kWh}$=200/1000×10=2 kWh.

Total per day $=1.5+2+2=5.5\text{ kWh}$=1.5+2+2=5.5 kWh.

Monthly energy $=5.5\times30=165\text{ kWh}=165\text{ units}$=5.5×30=165 kWh=165 units.

Cost $=165\times6=\text{Rs }990.$=165×6=Rs 990.

(a) The live wire enters through the main fuse and main switch; appliances are connected in parallel between live and neutral, with metal bodies joined to earth.

(b) Earthing: If insulation fails, the live wire may touch the metal body, raising it to a high potential. The earth wire, being of very low resistance, carries this large leakage current to the ground; this heavy current blows the fuse and disconnects the supply. Without earthing, a person touching the body would receive a fatal shock.

(c) Advantages of MCB over fuse:

1. An MCB can be reset by simply switching it on again; a blown fuse must be replaced.
2. An MCB responds faster and more precisely to overload and short-circuit, giving better protection.

Given: $P=2200\text{ W}$P=2200 W, $V=220\text{ V}$V=220 V.

(i) Current:
$I=\dfrac{P}{V}=\dfrac{2200}{220}=10\text{ A}.$I=P/V=2200/220=10 A.

(ii) Fuse rating: The fuse must be rated just above the normal current of $10\text{ A}$10 A. A $5\text{ A}$5 A fuse would blow at once and a $10\text{ A}$10 A fuse gives no margin, so the $15\text{ A}$15 A fuse is chosen.

(iii) Second motor: Two motors would draw $10+10=20\text{ A}$10+10=20 A, which exceeds the safe $12\text{ A}$12 A capacity of the wiring. So a second identical motor is not allowed on the same line; the wiring would overheat.

(iv) Energy in 30 min:
$E=P\times t=2200\times(30\times60)=2200\times1800=3.96\times10^{6}\text{ J}=3.96\text{ MJ}.$E=P× t=2200×(30×60)=2200×1800=3.96×10^6 J=3.96 MJ.

(i) One unit $=1\text{ kWh}=3.6\times10^{6}\text{ J}$=1 kWh=3.6×10^6 J.

(ii) Air-conditioner:
$E_{AC}=1.5\times8=12\text{ kWh}.$E_AC=1.5×8=12 kWh.

(iii) Total daily energy:
TV $=\dfrac{150}{1000}\times6=0.9\text{ kWh}$=150/1000×6=0.9 kWh.
Fans $=\dfrac{4\times75}{1000}\times10=\dfrac{300}{1000}\times10=3\text{ kWh}$=4×75/1000×10=300/1000×10=3 kWh.
$E_{\text{total}}=12+0.9+3=15.9\text{ kWh}.$E_total=12+0.9+3=15.9 kWh.

(iv) Monthly bill:
Monthly energy $=15.9\times30=477\text{ units}$=15.9×30=477 units.
$\text{Cost}=477\times7.50=\text{Rs }3577.50.$Cost=477×7.50=Rs 3577.50.