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Magnetic Effects of Electric Current - DOMESTIC ELECTRIC CIRCUITS12.4 DOMESTIC ELECTRIC CIRCUITS12.4 DOMESTIC ELECTRIC CIRCUITS12.4 DOMESTIC ELECTRIC CIRCUITS12.4 DOMESTIC ELECTRIC CIRCUITS

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Domestic power supply usually involves three wires: the Live wire (red insulation, positive), the Neutral wire (black insulation, negative), and the Earth wire (green insulation). In India, the potential difference between the live and neutral wires is 220 V220\text{ V}.

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All appliances in a domestic circuit are connected in parallel. This ensures that every appliance receives the full voltage of 220 V220\text{ V} and can be operated independently using its own switch.

Circuit diagram showing a lamp and a resistor connected in parallel between Live and Neutral lines.
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The Earth wire is a safety measure connected to a metal plate deep in the earth. It provides a low-resistance path for current, ensuring that any leakage to the metallic body of an appliance does not give a severe shock to the user.

Diagram of an appliance casing connected to an Earth ground for safety.
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Short-circuiting occurs when the Live wire and Neutral wire come into direct contact, causing the resistance of the circuit to become almost zero and the current to increase abruptly.

Diagram showing a direct vertical connection between Live and Neutral wires representing a short circuit.
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An Electric Fuse is a safety device made of a material with a low melting point. It is always connected in series with the live wire to protect the circuit from overcurrent due to overloading or short-circuiting.

Circuit showing a fuse in series with a load.

📐Formulae

P=V×IP = V \times I

I=PVI = \frac{P}{V}

H=I2RtH = I^2 R t

Rparallel=(1R1+1R2+⋯+1Rn)−1R_{parallel} = \left( \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n} \right)^{-1}

💡Examples

Problem 1:

An electric oven of 2 kW2\text{ kW} power rating is operated in a domestic electric circuit (220 V220\text{ V}) that has a current rating of 5 A5\text{ A}. What result do you expect? Explain.

Solution:

Given: P=2 kW=2000 WP = 2\text{ kW} = 2000\text{ W}, V=220 VV = 220\text{ V}. Current drawn I=PV=2000220≈9.09 AI = \frac{P}{V} = \frac{2000}{220} \approx 9.09\text{ A}.

Explanation:

The current drawn by the oven is 9.09 A9.09\text{ A}, which exceeds the circuit's current rating of 5 A5\text{ A}. This will lead to overloading, causing the fuse to melt and break the circuit to prevent potential fire hazards.

Problem 2:

Calculate the total current drawn from a 220 V220\text{ V} line if three appliances—a 100 W100\text{ W} bulb, a 1100 W1100\text{ W} heater, and a 500 W500\text{ W} refrigerator—are used simultaneously.

Solution:

Total Power Ptotal=100+1100+500=1700 WP_{total} = 100 + 1100 + 500 = 1700\text{ W}. Total Current I=PtotalV=1700220≈7.73 AI = \frac{P_{total}}{V} = \frac{1700}{220} \approx 7.73\text{ A}.

Explanation:

Since the appliances are in parallel, their powers are additive. The total current is found by dividing the total power by the supply voltage. A fuse of at least 8 A8\text{ A} or 10 A10\text{ A} would be required for this circuit.

Problem 3:

In a house, a 220 V220\text{ V} line is protected by a 5 A5\text{ A} fuse. How many 60 W60\text{ W} bulbs can be safely operated simultaneously in parallel without blowing the fuse?

Circuit diagram showing multiple 60W bulbs connected in parallel with a 5A fuse.

Solution:

Let nn be the number of bulbs. Total power P=n×60 WP = n \times 60\text{ W} Maximum current I=5 AI = 5\text{ A} Voltage V=220 VV = 220\text{ V} Pmax=V×IP_{max} = V \times I n×60=220×5n \times 60 = 220 \times 5 n×60=1100n \times 60 = 1100 n=110060≈18.33n = \frac{1100}{60} \approx 18.33 Therefore, a maximum of 1818 bulbs can be used.

Explanation:

To avoid blowing the fuse, the total current drawn by all bulbs must not exceed the fuse rating. Since the bulbs are in parallel, the total power is the product of the number of bulbs and the power of one bulb. We solve for nn and round down to the nearest whole number to ensure the current stays below the 5 A5\text{ A} limit.

Problem 4:

An electric iron with a power rating of 1500 W1500\text{ W} is connected to a 220 V220\text{ V} domestic circuit. If the circuit is protected by a 5 A5\text{ A} fuse, determine whether the fuse will blow when the iron is switched on. Support your answer with calculations.

A circuit diagram showing a 220V power source, a 5A fuse, and an electric iron load connected in series.

Solution:

P=1500 WP = 1500\text{ W} V=220 VV = 220\text{ V} I=PV=1500220≈6.82 AI = \frac{P}{V} = \frac{1500}{220} \approx 6.82\text{ A} Since 6.82 A>5 A6.82\text{ A} > 5\text{ A}, the fuse will blow.

Explanation:

The current drawn by the electric iron is calculated using the formula I=PVI = \frac{P}{V}. Substituting the given values, we find the current to be approximately 6.82 A6.82\text{ A}. Since the fuse in the circuit has a maximum capacity of 5 A5\text{ A}, the current exceeds this limit, causing the fuse wire to melt and break the circuit to prevent damage and fire hazards.