Electrical Safety

Question

Step-by-step solution

1. Step 1

   Find the operating current.

   $I = \dfrac{P}{V} = \dfrac{2300}{230} = 10\,\text{A}$

2. Step 2

   Pick the fuse just ABOVE the operating current.

Question

Step-by-step solution

1. Step 1

   If a fault makes the casing live, the earth wire provides a low-resistance path to ground.

2. Step 2

   A large current flows → blows the fuse → cuts the supply, preventing electric shock.

Question

Step-by-step solution

1. Step 1

   Plastic casing is an INSULATOR — no metal can become live.

2. Step 2

   Two layers of insulation protect the user; earthing is unnecessary.

Question

Step-by-step solution

1. Step 1

   Power $P = I^{2}R$ → heating in the cable rises with the SQUARE of current.

2. Step 2

   If the appliances draw more current than the cable is rated for, insulation overheats and may ignite.

Question

Step-by-step solution

1. Step 1

   $P = VI \Rightarrow I = P/V$.

   $I = 1200/230 \approx 5.2\,\text{A}$

Question

Step-by-step solution

1. Step 1

   Convert power to kilowatts and time to hours.

   $P = 2.0\,\text{kW};\quad t = 9/60 = 0.15\,\text{h}$

2. Step 2

   Energy in kWh = $P \times t$.

   $E = 2.0 \times 0.15 = 0.30\,\text{kWh}$

Question

Step-by-step solution

1. Step 1

   Energy per day: $E = P \times t$.

   $E_{\text{day}} = 2.5 \times 4 = 10\,\text{kWh}$

2. Step 2

   Energy in 30 days.

   $E_{30} = 10 \times 30 = 300\,\text{kWh}$

3. Step 3

   Cost = energy × unit price.

   $C = 300 \times 30 = 9000\,\text{cents} = \$90$

Question

Step-by-step solution

1. Step 1

   The fuse only blows when current EXCEEDS its rating ($13\,\text{A}$). $30\,\text{mA} = 0.03\,\text{A}$ is far below the rating — the fuse does not melt.

2. Step 2

   An RCD continuously compares current in the LIVE wire with current returning in the NEUTRAL wire.

3. Step 3

   If current leaks through the person to earth, the live and neutral currents differ by that amount. When the imbalance exceeds the $30\,\text{mA}$ trip rating, the RCD opens the circuit in $\sim 30\,\text{ms}$ — fast enough to prevent serious electric shock.

Question

Step-by-step solution

1. Step 1

   Find the operating current.

   $I = \dfrac{P}{V} = \dfrac{920}{230} = 4.0\,\text{A}$

2. Step 2

   Choose the fuse just ABOVE the operating current.

3. Step 3

   $3\,\text{A}$ would blow under normal use. $13\,\text{A}$ is far too high — a fault current of $\sim 10\,\text{A}$ would not blow it. $5\,\text{A}$ is just above $4\,\text{A}$ and gives good fault protection.

Question

Step-by-step solution

1. Step 1

   Metal case + earth: if a fault makes the case live, fault current flows to earth, blows the fuse, isolates the supply. BUT relies on the earth wire being intact AND the fuse blowing quickly.

2. Step 2

   Plastic case + double insulation: the casing itself is an insulator — there is no metal part that the user can touch that could become live, even after a single insulation failure.

3. Step 3

   Double insulation does not require an earth wire (often marked with a 'square inside a square' symbol). It is the preferred approach for hand-held appliances where the user grips the case continuously.