Question

A $1\,kW$ heater is meant to operate at $200\,V$. How much power (in watt) will it consume if the line voltage drops to $100\,V$ ?

Answer 1

The electrical resistance of an object is a measure of its resistance to the flow of electric current in electronics and electromagnetism. Electrical conductance is the ease with which an electric current travels in the reciprocal quantity.

Formula used:
$VI = P$
Here, $P$ = Power, $V$ = Potential difference and $I$ = Current.

Complete step by step answer:
The quantity of energy moved or transformed per unit time is referred to as power in physics. The watt, which is equal to one joule per second in the International System of Units, is the unit of power. Power is also referred to as activity in ancient writings. A scalar quantity is power.

The rate at which energy is transformed from the electrical energy of moving charges to another form, such as heat, mechanical energy, or energy stored in electric fields or magnetic fields, is represented by the electric power in watts associated with a whole electric circuit or a circuit component. The product of applied voltage and electric current gives the power of a resistor in a DC circuit:
$VI = P$

We know that according to Ohm's law. If all physical parameters and temperature stay constant, Ohm's law asserts that the voltage across a conductor is precisely proportional to the current flowing through it.
$V = IR$
$\Rightarrow I = \dfrac{V}{R}$
Now rearranging the two formulas we get,
$P = \dfrac{{{V^2}}}{R}$
Now a 1kW heater is meant to operate at 200 V.
$P = 1000\,W \\\ \Rightarrow V = 200\,V \\\$
$\Rightarrow R = \dfrac{{{V^2}}}{P} \\\$
$\Rightarrow {\mathbf{R}} = \dfrac{{{{\mathbf{V}}^2}}}{{\mathbf{P}}} \\\ \Rightarrow R= \dfrac{{{\mathbf{20}}{{\mathbf{0}}^2}}}{{{\mathbf{1000}}}} \\\ \Rightarrow R= 40\Omega$

Now we find how much power (in watt) will it consume if the line voltage drops to 100V?
$V = 100\,V$
$\Rightarrow R = 40\Omega$
Using
$P = \dfrac{{{V^2}}}{R}$
$\Rightarrow {\mathbf{P}} = \dfrac{{{{\mathbf{V}}^2}}}{{\mathbf{R}}} \\\ \Rightarrow P = \dfrac{{{\mathbf{10}}{{\mathbf{0}}^2}}}{{{\mathbf{40}}}} \\\$\therefore P = 25,W$

Hence, a heater consumes a power of 25 W.

Note: Ohm's law is only valid if the temperature and other physical variables stay constant. Increasing the current in some components boosts the temperature. The filament of a light bulb, for example, increases in temperature when the current is raised. Ohm's law does not apply in this situation. Ohm's Law is broken by the lightbulb filament.