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Question: An electric lamp is marked \[60{\text{ W}}\], \[{\text{240 V}}\]. What is the current passing throug...

An electric lamp is marked 60 W60{\text{ W}}, 240 V{\text{240 V}}. What is the current passing through it when connected to 220 V{\text{220 V}} power supply? How many joules does it consume in1 hr1{\text{ }}hr?

Explanation

Solution

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 W60{\text{ W}}, 240 V{\text{240 V}}, it means that the consumes 60 J60{\text{ J}} of energy per unit time when connected to a 240 V{\text{240 V}} power supply.

Formula used:
The power of the electric lamp is associated with the voltage applied to it as
P=V2RP = \dfrac{{{V^2}}}{R} …… (1)
Here, PP is the power of the lamp, VV is the voltage applied, and RR is the resistance of the electric bulb.
And, the Ohm's Law is given by
V=IRV = IR …… (2), where II 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 W60{\text{ W}} and the marked voltage is 240 V{\text{240 V}}
Substitute 60 W60{\text{ W}} for PP, and 240 V{\text{240 V}} for VV in equation (1) as
60=2402R60 = \dfrac{{{{240}^2}}}{R}
R=240×24060\Rightarrow R = \dfrac{{240 \times 240}}{{60}}
R=960 Ω\Rightarrow R = 960{\text{ }}\Omega
Substitute 960 Ω960{\text{ }}\Omega for RR, and 220 V{\text{220 V}} for VV in equation (2) as
220=I(960)220 = I\left( {960} \right)
I=220960\Rightarrow I = \dfrac{{220}}{{960}}
I=0.23 A\Rightarrow I = 0.23{\text{ A}}
Therefore, the current passing through the electric lamp is 0.23 A0.23{\text{ A}}.
Now, substitute 220 V{\text{220 V}}for VV, and 960 Ω960{\text{ }}\Omega for RR in equation (1) as
P=2202960P = \dfrac{{{{220}^2}}}{{960}}
P=48400960\Rightarrow P = \dfrac{{48400}}{{960}}
P=50.42W\Rightarrow P = 50.42{\text{W}}
Also, the power is given defined by work done per unit time and is expressed as
P=WtP = \dfrac{W}{t} …… (3)
Here, WW is the work done and tt is the time.
Substitute 50.42 W50.42{\text{ W}} for PP and 1 hr1{\text{ hr}} for tt in equation (3) as
50.42 W=W1 hr50.42{\text{ W}} = \dfrac{W}{{1{\text{ hr}}}}
W=50.42 W×1 hr\Rightarrow W = 50.42{\text{ W}} \times 1{\text{ hr}}
W=50.42 W×3600 s\Rightarrow W = 50.42{\text{ W}} \times 3600{\text{ s}}
W = 181512 J\Rightarrow {\text{W = 181512 J}}
Therefore, the electric lamp consumes 181512 J{\text{181512 J}} in 1 hr1{\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, W{\text{W}}, which is equal to one joule of energy consumed in one second.