Calculate the current through a $60\,W$ lamp rated $250\,V$. If the line voltage falls to $200\,V$, how is Power consumed by the lamp affected ?

Hint: From the definition of power of an electric circuit find the power of the lamp for both the voltages and compare them. The power of an electric circuit is equal to the product of the voltage and the current through the circuit.

Formula used:
The power of an electric circuit in terms of voltage and resistance is given by,
\[P = \dfrac{{{V^2}}}{R}\]
where, \[V\] is the potential difference between the terminals of the circuit and \[R\] is the resistance of the circuit.

Complete step by step answer:
We have given here a lamp of 60 W which is rated at 250V. So, we have to find the resistance of the lamp first. Now, we know that the power of a circuit element is given by,
\[P = \dfrac{{{V^2}}}{R}\]
Here, we have given, \[P = 60\,W\] and \[V = 250\,V\]. So, putting the values we will have,
\[60 = \dfrac{{{{250}^2}}}{R}\]
\[\Rightarrow R = \dfrac{{{{250}^2}}}{{60}}\]
\[\Rightarrow R = 1041.67\]
So, the resistance of the lamp is \[1041.67\Omega \]
Now, if the voltage drops to \[V = 200V\] the power will be,
\[P = \dfrac{{{{200}^2}}}{{1041.67}} \\
\therefore P= 38.39\,W\]

Hence, the power dissipation of the lamp will decrease to \[38.39\,W\] when the voltage drops to \[200\,V\].

Note: Remember that the resistance of any circuit component like the bulb or fan is constant, but the voltage and current varies as per the source, the maximum limit only depends on the value of the power. So, when solving this type of problem we have to keep in mind that the only constant of the equation is the power and the resistance of the lamp, so we have written the formula for the power in terms of voltage and resistance only.