
An electric bulb is marked $100\,W,230\,V$ if the supply voltage drops to $115\,V$ what will be the total heat and light energy, totally liberated by the bulb in $20$ minutes?
Answer
512.4k+ views
Hint: In order to solve this question we need to understand electric current and production of heat using current. Electric current is defined as total charge crossing per unit area. Actually when the electrical circuit is subjected to potential difference then an electrical field is set up in wire which binds the electrons to move in opposite direction as applied electrical field which in turn allows electric current to flow in direction of electric field or in direction of decreasing potential. When an electric current flows inside wire then wire opposes this behavior and generates resistance which dissipates the electrical energy to heat energy.
Complete step by step answer:
Given the power of the bulb rated, $P = 100\,W$.
And the rated voltage across the bulb is, ${V_1} = 230\,V$.
Let the resistance of the bulb be, $R$.
Then from power of bulb,
$P = \dfrac{{{V_1}^2}}{R}$
So resistance of bulb is,
$R = \dfrac{{{V_1}^2}}{P}$
Putting values we get,
$R = \dfrac{{{{(230)}^2}}}{{100}}\Omega $
$\Rightarrow R = 529\Omega $
Now since the voltage dropped to, ${V_2} = 115V$
New power would be,
$P' = \dfrac{{{V_2}^2}}{R}$
Putting values we get,
$P' = \dfrac{{{{(115)}^2}}}{{529}}$
$\Rightarrow P' = 25\,W$
Time given in problem, $t = 20\min $
$t = 20 \times 60\sec $
$\Rightarrow t = 1200\sec $
Heat dissipated is, $H = P' \times t$
Putting values we get,
$H = 25 \times 1200\,Watt\sec $
$\Rightarrow H = 30000J$
$\therefore H = 30kJ$
So heat dissipated would be $30\,KJ$.
Note: It should be remembered that we have calculated using resistance but it can also be calculated using max current drawn by the bulb to glow and later solving it. Electric bulb is made up of tungsten filament, so when current passes through it, the tungsten filament heats up and hence the light energy travels but since the bulb is enclosed so the electric bulb glows.
Complete step by step answer:
Given the power of the bulb rated, $P = 100\,W$.
And the rated voltage across the bulb is, ${V_1} = 230\,V$.
Let the resistance of the bulb be, $R$.
Then from power of bulb,
$P = \dfrac{{{V_1}^2}}{R}$
So resistance of bulb is,
$R = \dfrac{{{V_1}^2}}{P}$
Putting values we get,
$R = \dfrac{{{{(230)}^2}}}{{100}}\Omega $
$\Rightarrow R = 529\Omega $
Now since the voltage dropped to, ${V_2} = 115V$
New power would be,
$P' = \dfrac{{{V_2}^2}}{R}$
Putting values we get,
$P' = \dfrac{{{{(115)}^2}}}{{529}}$
$\Rightarrow P' = 25\,W$
Time given in problem, $t = 20\min $
$t = 20 \times 60\sec $
$\Rightarrow t = 1200\sec $
Heat dissipated is, $H = P' \times t$
Putting values we get,
$H = 25 \times 1200\,Watt\sec $
$\Rightarrow H = 30000J$
$\therefore H = 30kJ$
So heat dissipated would be $30\,KJ$.
Note: It should be remembered that we have calculated using resistance but it can also be calculated using max current drawn by the bulb to glow and later solving it. Electric bulb is made up of tungsten filament, so when current passes through it, the tungsten filament heats up and hence the light energy travels but since the bulb is enclosed so the electric bulb glows.
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