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Physics: Electricity and Magnetism - Overloading, Short Circuits, Fuses, and Earthing

Grade 8IB

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

🔑Concepts

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Overloading occurs when too many high-power electrical appliances (like heaters and air conditioners) are connected to a single socket or circuit, causing the total current ItotalI_{total} to exceed the safety limit of the wires, leading to overheating.

Circuit diagram showing multiple resistors in parallel representing multiple appliances connected to the same voltage source, increasing total current.
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A short circuit happens when the live wire and neutral wire come into direct contact due to damaged insulation. This provides a path of nearly zero resistance (R≈0ΩR \approx 0 \Omega), causing current to surge to dangerous levels.

Circuit diagram showing a low-resistance path (short) bypassing the load between live and neutral wires.
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A fuse is a safety device containing a thin wire with a low melting point. It is connected in series with the live wire. If the current exceeds the fuse rating, the wire melts (H=I2RtH = I^2Rt), breaking the circuit.

Circuit diagram showing a fuse symbol in series with an appliance.
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Earthing (Grounding) involves connecting the metallic body of an appliance to the earth via a low-resistance wire. If a fault causes the live wire to touch the metallic casing, the current flows safely to the ground instead of through a person.

Diagram of an appliance body connected to a ground symbol.

📐Formulae

P=V×IP = V \times I

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

V=I×RV = I \times R

H=I2×R×tH = I^2 \times R \times t

💡Examples

Problem 1:

An electric oven with a power rating of 3000 W3000 \text{ W} is operated in a domestic circuit that has a voltage of 220 V220 \text{ V}. If the circuit is protected by a 10 A10 \text{ A} fuse, will the fuse blow when the oven is switched on?

Solution:

Given P=3000 WP = 3000 \text{ W} and V=220 VV = 220 \text{ V}. Using the formula I=PVI = \frac{P}{V}: I=3000220≈13.64 AI = \frac{3000}{220} \approx 13.64 \text{ A} Since 13.64 A>10 A13.64 \text{ A} > 10 \text{ A}, the fuse will melt and break the circuit.

Explanation:

The current required by the oven is higher than the fuse rating. This is a case of potential overloading, and the fuse acts as a safety measure to prevent damage.

Problem 2:

Calculate the resistance RR of a circuit during a short circuit if the voltage is 240 V240 \text{ V} and the current surges to 480 A480 \text{ A}.

Solution:

Using Ohm's Law V=I×RV = I \times R, we can rearrange it to find RR: R=VIR = \frac{V}{I} R=240480=0.5ΩR = \frac{240}{480} = 0.5 \Omega

Explanation:

During a short circuit, the resistance becomes very low (in this case, 0.5Ω0.5 \Omega), which allows a very high current to flow through the wires.

Problem 3:

A 2.2 kW2.2 \text{ kW} electric kettle is used on a 220 V220 \text{ V} supply. Determine the current flowing through the kettle and select an appropriate fuse from the following ratings: 3 A3 \text{ A}, 5 A5 \text{ A}, or 13 A13 \text{ A}.

Circuit with a 220V source, a 13A fuse, and a 2.2kW kettle load.

Solution:

  1. Convert power to Watts: P=2.2 kW=2200 WP = 2.2 \text{ kW} = 2200 \text{ W}
  2. Use the formula for current: I=PVI = \frac{P}{V} I=2200220I = \frac{2200}{220} I=10 AI = 10 \text{ A}
  3. The fuse rating must be slightly higher than the operating current. Therefore, a 13 A13 \text{ A} fuse is required.

Explanation:

A 5 A5 \text{ A} fuse would blow immediately under normal operation as 10 A>5 A10 \text{ A} > 5 \text{ A}. A 13 A13 \text{ A} fuse allows the normal 10 A10 \text{ A} current but will blow if a fault occurs.

Problem 4:

An electric iron has an internal resistance of 40Ω40 \Omega. A fault creates a short circuit path with a resistance of only 2Ω2 \Omega. Calculate the current surge in the circuit if the supply is 240 V240 \text{ V}.

Circuit diagram showing a normal 40 Ohm load and a dashed 2 Ohm short circuit path in parallel.

Solution:

  1. Under normal conditions: Inormal=VRiron=24040=6 AI_{normal} = \frac{V}{R_{iron}} = \frac{240}{40} = 6 \text{ A}
  2. Under fault conditions (short circuit): Ifault=VRfault=2402I_{fault} = \frac{V}{R_{fault}} = \frac{240}{2} Ifault=120 AI_{fault} = 120 \text{ A}
  3. The current increases from 6 A6 \text{ A} to 120 A120 \text{ A}.

Explanation:

The drastic decrease in resistance causes a massive surge in current. Since heating effect H∝I2H \propto I^2, the heat generated increases by a factor of (20)2=400(20)^2 = 400, which can melt wires and cause a fire.