How many `60 W` lamps may be safely run on a `230 V` circuit fitted with a ` 5 A` fuse?

Options

A. `19
B. `22`
C. `33`
D. `66`

Explanation

To determine how many 60 W lamps can be safely run on a 230 V circuit fitted with a 5 A fuse, we can follow these steps:

### Step 1: Calculate the current drawn by one lamp
The power (P) of one lamp is given as 60 W, and the voltage (V) is 230 V. We can use the formula for power:

\[ P = V \times I \]

Rearranging this formula to find the current (I) gives us:

\[ I = \frac{P}{V} \]

Substituting the values:

\[ I = \frac{60 \, \text{W}}{230 \, \text{V}} = \frac{60}{230} \, \text{A} \approx 0.26087 \, \text{A} \]

### Step 2: Determine the total current for n lamps
If we have n lamps connected in parallel, the total current (I_total) drawn by n lamps can be expressed as:

\[ I_{\text{total}} = n \times I \]

Substituting the expression for I:

\[ I_{\text{total}} = n \times \frac{60}{230} \]

### Step 3: Set up the inequality based on the fuse rating
The fuse is rated for 5 A, which means the total current must not exceed this value:

\[ I_{\text{total}} \leq 5 \, \text{A} \]

Substituting our expression for I_total:

\[ n \times \frac{60}{230} \leq 5 \]

### Step 4: Solve for n
To find n, we can rearrange the inequality:

\[ n \leq 5 \times \frac{230}{60} \]

Calculating the right side:

\[ n \leq \frac{1150}{60} \approx 19.1667 \]

Since n must be a whole number, we take the largest whole number less than or equal to 19.1667, which is 19.

### Conclusion
Thus, the maximum number of 60 W lamps that can be safely run on a 230 V circuit fitted with a 5 A fuse is:

\[ \boxed{19} \]

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