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An electric lamp is marked \[60W,240V\]. What is the current passing through it when connected to a \[220V\] power supply? How many joules does it consume in \[1hr\] ?

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Last updated date: 23rd May 2024
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Answer
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Hint: ‘The electric lamp is marked’ means there some readings of the voltage and power of the lamp are given. The resistance of the lamp has to be calculated from the given power and voltage. By applying, Ohm’s law the current can be calculated from the given voltage and calculated resistance of the lamp. Note that the resistance is a constant quantity. Hence there should be no change in the resistance for changing the voltage of the power supply.
Formula used:
For a power reading \[P\] , voltage of the power supply $V$ and resistance of the lamp $R$
\[P = \dfrac{{{V^2}}}{R}\]
From ohm’s law, \[V = IR\], $I$ is the current passing through the lamp.

Complete step by step answer:
The lamp is marked with two readings. One is power and another is the voltage of the power supply.
Here given,
power reading\[P = 60W\]
the voltage of the power supply $V = 240V$
the equation of the power in terms of the voltage supply is, \[P = \dfrac{{{V^2}}}{R}\]
resistance of the lamp is $R$.
\[\therefore R = \dfrac{{{V^2}}}{P}\]
\[
  R = \dfrac{{240 \times 240}}{{60}} \\
  R = 960\Omega \\
\]
Now, applying Ohm’s law we get \[V = IR\], $I$ is the current passing through the lamp.
$ \Rightarrow I = \dfrac{V}{R}$
Given that the lamp is connected to a voltage $V = 220V$
$ \Rightarrow I = \dfrac{{220}}{{960}}$
$ \Rightarrow I = 0.229A \simeq 0.23A$
Hence, the current passing through the lamp is $0.23A$ when connected to a \[220V\] power supply.
Now, from Joule’s heat equation we get, consumed heat in time $t$, $H = {I^2}Rt$
Given, $t = 1hr = 3600\sec $ and we get from the above calculations, $I = 0.23A,R = 960\Omega $
$\therefore H = 0.23 \times 0.23 \times 960 \times 3600$
$ \Rightarrow H = 182822.4J$
Hence, it consumes $182822.4Joules$ in \[1hr\].

Note: An electric lamp is a standard light-emitting element employed in totally different circuits, principally for lighting and indicating functions. The development of the lamp is kind of easy, it's one filament close to that, a clear glass created spherical cover is provided. The filament of the lamp is the principal product of metal because it has a high freezing point temperature. A lamp emits light energy because the thin little metal filament of the lamp glows while not being liquid, whereas current flows through it.
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