Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 21st Jul 2024
Total views: 349.2k
Views today: 8.49k
Answer
VerifiedVerified
349.2k+ views
Hint: The current passing through the electric lamp can be calculated by applying Ohm’s law to it which states that the current passing through the lamp is directly proportional to the voltage applied across it. When an electric lamp is marked as \[60{\text{ W}}\], \[{\text{240 V}}\], it means that the consumes \[60{\text{ J}}\] of energy per unit time when connected to a \[{\text{240 V}}\] power supply.

Formula used:
The power of the electric lamp is associated with the voltage applied to it as
\[P = \dfrac{{{V^2}}}{R}\] …… (1)
Here, \[P\] is the power of the lamp, \[V\] is the voltage applied, and \[R\] is the resistance of the electric bulb.
And, the Ohm's Law is given by
\[V = IR\] …… (2), where \[I\] is the current passing through the circuit.

Complete step by step answer:
In the question, it is given that the power of the electric lamp is \[60{\text{ W}}\] and the marked voltage is \[{\text{240 V}}\]
Substitute \[60{\text{ W}}\] for \[P\], and \[{\text{240 V}}\] for \[V\] in equation (1) as
\[60 = \dfrac{{{{240}^2}}}{R}\]
\[ \Rightarrow R = \dfrac{{240 \times 240}}{{60}}\]
\[ \Rightarrow R = 960{\text{ }}\Omega \]
Substitute \[960{\text{ }}\Omega \] for \[R\], and \[{\text{220 V}}\] for \[V\] in equation (2) as
\[220 = I\left( {960} \right)\]
\[ \Rightarrow I = \dfrac{{220}}{{960}}\]
\[ \Rightarrow I = 0.23{\text{ A}}\]
Therefore, the current passing through the electric lamp is \[0.23{\text{ A}}\].
Now, substitute \[{\text{220 V}}\]for \[V\], and \[960{\text{ }}\Omega \] for \[R\] in equation (1) as
\[P = \dfrac{{{{220}^2}}}{{960}}\]
\[ \Rightarrow P = \dfrac{{48400}}{{960}}\]
\[ \Rightarrow P = 50.42{\text{W}}\]
Also, the power is given defined by work done per unit time and is expressed as
\[P = \dfrac{W}{t}\] …… (3)
Here, \[W\] is the work done and \[t\] is the time.
Substitute \[50.42{\text{ W}}\] for \[P\] and \[1{\text{ hr}}\] for \[t\] in equation (3) as
\[50.42{\text{ W}} = \dfrac{W}{{1{\text{ hr}}}}\]
\[ \Rightarrow W = 50.42{\text{ W}} \times 1{\text{ hr}}\]
\[ \Rightarrow W = 50.42{\text{ W}} \times 3600{\text{ s}}\]
\[ \Rightarrow {\text{W = 181512 J}}\]
Therefore, the electric lamp consumes \[{\text{181512 J}}\] in \[1{\text{ hr}}\].

Note: Power defines the amount of energy that is converted or transferred in an electric circuit per unit of time. It's S.I. The unit is Watt, \[{\text{W}}\], which is equal to one joule of energy consumed in one second.