The mains power supply of a house is through a $5\,A$ fuse. How many $100 W,220V$ bulbs can be used in this house at the correct voltage?
Answer
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Hint:To answer this question, we first need to understand what a fuse is.Fuses are sacrificial devices that are used to protect much more expensive electrical components from overcurrent damage. They are made up of a low-resistance metal or wire used to complete a circuit.
Complete step by step answer:
Working of a fuse: An electric fuse is a safety device that consists of a piece of wire having a low melting point. When a circuit is subjected to excessive current, it melts and breaks when the temperature exceeds its melting point. It is used to prevent short circuiting and so safeguard electrical gadgets from harm.
As given in the question
${I_{\max }} = 5A{\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} V = 220V{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {\kern 1pt} {P_B} = 100W$
As we know that formula of power is - $P = \dfrac{{{V^2}}}{R}$.
putting the values in this equation,
$100W = \dfrac{{{{220}^2}}}{R}$........(we take voltage as 220V because in a house bulb are always connected in parallel)
By solving this we get the value of $R$ as
$R = \dfrac{{220 \times 220}}{{100}} = 484\Omega $..........(value of resistance of each bulb)
$\Rightarrow {R_r} = \dfrac{V}{{{I_{\max }}}}$...........(here ${R_r}$is the required value of resistance)
$\Rightarrow {R_r} = \dfrac{{220}}{5} = 44\Omega $
Now in parallel connection when all the elements are of same resistance than net resistance is given as -${R_{net}} = \dfrac{R}{n}$ (here n is the number of elements and R is the resistance of each element)
Substituting the values
$44 = \dfrac{{484}}{n}$
By solving we get $n$ as
$\therefore n = 11$
So, the total number of bulbs required is 11 bulbs.
Note: Both fuses and circuit breakers have the same purpose: to protect electrical circuits from overloads that can cause fires. They both stop the flow of electricity, but they do so in quite different ways.
Complete step by step answer:
Working of a fuse: An electric fuse is a safety device that consists of a piece of wire having a low melting point. When a circuit is subjected to excessive current, it melts and breaks when the temperature exceeds its melting point. It is used to prevent short circuiting and so safeguard electrical gadgets from harm.
As given in the question
${I_{\max }} = 5A{\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} V = 220V{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {\kern 1pt} {P_B} = 100W$
As we know that formula of power is - $P = \dfrac{{{V^2}}}{R}$.
putting the values in this equation,
$100W = \dfrac{{{{220}^2}}}{R}$........(we take voltage as 220V because in a house bulb are always connected in parallel)
By solving this we get the value of $R$ as
$R = \dfrac{{220 \times 220}}{{100}} = 484\Omega $..........(value of resistance of each bulb)
$\Rightarrow {R_r} = \dfrac{V}{{{I_{\max }}}}$...........(here ${R_r}$is the required value of resistance)
$\Rightarrow {R_r} = \dfrac{{220}}{5} = 44\Omega $
Now in parallel connection when all the elements are of same resistance than net resistance is given as -${R_{net}} = \dfrac{R}{n}$ (here n is the number of elements and R is the resistance of each element)
Substituting the values
$44 = \dfrac{{484}}{n}$
By solving we get $n$ as
$\therefore n = 11$
So, the total number of bulbs required is 11 bulbs.
Note: Both fuses and circuit breakers have the same purpose: to protect electrical circuits from overloads that can cause fires. They both stop the flow of electricity, but they do so in quite different ways.
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