Question

The resistance of the filament of a lamp increases with the increase in temperature. A lamp rated $100\,W$ and $220\,V$ is connected across $220\,V$ power supply. If the voltage drops by $10\%$, then the power of the lamp be:
A. $90\,W$
B. $81\,W$
C. between $90\,W$ and $100\,W$
D. between $81\,W$ and $90\,W$

Answer 1

Hint- We know that power is the product of current and voltage.
$P = I \times V$
Where, $I$ is the current and $V$ is the voltage
From ohm’s law, we know
$V = IR \\\ I = \dfrac{V}{R} \\\$
Therefore, $P = \dfrac{{{V^2}}}{R}$
From this, resistance is
$R = \dfrac{{{V^2}}}{P}$
Let $R$ be a constant resistance. Then, Power when voltage decreased by $10\%$ is given as,
$P = \dfrac{{{{\left( {V - \dfrac{{10}}{{100}}V} \right)}^2}}}{R}$
Let $R'$be the resistance changing with temperature. If voltage in a circuit decreases then current flowing in the circuit decreases. Hence the heating of the resistor will be less. So, this new resistance $R'$ will be less than $R$
Power in this case is
$P = \dfrac{{{{\left( {V - \dfrac{{10}}{{100}}V} \right)}^2}}}{{{R’}}}$

Step by step solution:
Given,
Power, $P = 100\,W$
Voltage, $V = 220\,V$
We know that power is the product of current and voltage.
$P = I \times V$
Where, $I$ is the current and $V$ is the voltage.
From ohm’s law, we know
$V = IR \\\ I = \dfrac{V}{R} \\\$
Substituting this in equation (1) we get,
$P = \dfrac{{{V^2}}}{R}$
From this, resistance is
$R = \dfrac{{{V^2}}}{P}$
On substituting the given values, we get,
$R = {\dfrac{{\left( {220} \right)}}{{100}}^2} = 484\,\Omega$
Suppose this resistance $R$ remain same when voltage drops
Initial voltage $= 220\,V$
Final voltage= $V - \dfrac{{10}}{{100}}V = 220 - \dfrac{{10 \times 220}}{{100}} = 198\,V$
Therefore,
Power when voltage decreased by $10\%$ is given as,
$P = \dfrac{{{{\left( {V - \dfrac{{10}}{{100}}V} \right)}^2}}}{R}$
$P = {\dfrac{{\left( {198} \right)}}{{484}}^2}\, = 81\,W$
But in our question resistance changes with temperature. Let $R'$be the resistance changing with temperature.
If voltage in a circuit decreases then current flowing in the circuit decreases. Hence the heating of the resistor will be less. So, this new resistance $R'$ will be less than $R$
Now, power in this case is $P = \dfrac{{{{\left( {V - \dfrac{{10}}{{100}}V} \right)}^2}}}{R’}$
= ${\dfrac{{\left( {198} \right)}}{{{R'}}}^2}$
Since $R'$ is less than $R$ power here will be more than that for resistance $R$.
For constant resistance we got power as $81\,W$
So, for $R'$ power will be more than this value.

So, the correct option is option D.

Note: In our question resistance changes with temperature. Let $R'$be the resistance changing with temperature. According to ohm’s law if voltage in a circuit decreases then current flowing in the circuit decreases. Hence the heating of the resistor will be less. It is given that the resistance increases with temperature. Since the heating is less in this case the new resistance will be less than the initial value.