Question

An electric geyser has the ratings 2000 W, 220 V marked on it. What should be the minimum rating, in whole number, of a fuse wire that may be required for safe use with this geyser?

Answer 1

Hint:In the given question, we are given the ratings of an electric geyser and we have to calculate the minimum rating of the fuse for safe usage. So, we will first find out the current that will flow through the geyser with the ratings as specified in the question. Then, the same current will flow through the fuse wire also. So, we will round up the current to the next whole number for safe usage.

Complete step by step answer:
In the question, we are given the power and voltage rating of an electric geyser. So, we have,
$P=2000\,W$
$\Rightarrow V=220\,V$
We have to find the minimum setting of a fuse wire for safety. As we know, the electric power generated in the circuit is the energy produced in the resistor per unit time. Thus, we have,
$P=VI$
Shifting the terms in the equation, we have,
$I=\dfrac{P}{V}$
Substituting the values of known quantities, we get,
$I=\dfrac{2000}{220}$
$\therefore I=9.09A$
Fuse should always have its rating higher than the flowing current so that it can melt and open the circuit if excessive current will flow and avoid fire.

Hence, the minimum whole number rating is $10A$ required for safe use of the geyser.

Note:If the resistances are connected in series, then using $P={{I}^{2}}R$, the power developed will be higher in the resistor of higher value as current will be the same in all resistors. If resistances are connected in parallel, then using $P=\dfrac{{{V}^{2}}}{R}$ , the power developed will be higher in the resistor of lower value as potential will be same across all resistors.Usually, filament bulbs get fused when they are switched on. This is because with the rise in temperature, the resistance of the bulb increases and becomes constant in steady state.