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# Two $220V$, $100W$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220V$ AC supply line, the power drawn by the combination in each case respectively will be\begin{align} & A)50W,100W \\ & B)100W,50W \\ & C)200W,150W \\ & D)50W,200W \\ \end{align}

Hint: Here we will use the relation between power, voltage and resistance to determine the resistance of an individual bulb. Power is given as the ratio of square of the voltage limit of the bulb to the resistance. Then we will find the equivalent resistance of the bulb when it is connected in series connection and parallel connection and use the same relation between power, voltage and resistance to determine the power in each combination of bulbs. We need to use the source voltage and equivalent resistance in this case.

Formula used:
$P=\dfrac{{{V}^{2}}}{R}$

Firstly, we will find the resistance offered by a single bulb by using the voltage limit and power limit of the bulb which are given as $220V$ and $100W$ respectively. The resistance of a single bulb will be,
$R=\dfrac{{{V}^{2}}}{P}=\dfrac{{{\left( 220 \right)}^{2}}}{100}=484\Omega$
Now, if we connect two of these bulbs in a series connection, the equivalent resistance will be,
${{R}_{\text{equivalent}}}=R+R=2R=2\times 484=968\Omega$
So, we have the equivalent resistance when the bulbs are connected in series. Now, to determine the power when connected to a $220V$ source, we will use the expression,
$P=\dfrac{{{V}^{2}}}{{{R}_{\text{equivalent}}}}=\dfrac{{{\left( 220 \right)}^{2}}}{968}=50W$
So, the power drawn by the series combination of these bulbs is $50W$.
Now, if we connect these two bulbs in parallel connection, the equivalent resistance will be,
\begin{align} & \dfrac{1}{{{R}_{\text{equivalent}}}}=\dfrac{1}{R}+\dfrac{1}{R}=\dfrac{2}{R}=\dfrac{2}{484}=\dfrac{1}{242} \\ & \Rightarrow {{R}_{\text{equivalent}}}=242\Omega \\ \end{align}
So, we have equivalent resistance when the given bulbs are connected in parallel. Now, the power drawn while connecting these bulbs in parallel is,
$P=\dfrac{{{V}^{2}}}{{{R}_{\text{equivalent}}}}=\dfrac{{{\left( 220 \right)}^{2}}}{242}=100W$
So, the power drawn by the parallel combination of these bulbs is $100W$.
Therefore, we can conclude that the power drawn when two of the given bulbs are connected in series combination is $50W$ and in parallel combination is $100W$.

So, option a is the correct answer.

Note:
We can also solve this question by using the relation between power, current and resistance which is given as $P={{I}^{2}}R$. But instead of using the given voltage supply directly, we will need to find the current passing through the circuit by using ohm's law. While finding this current we must remember to use the equivalent resistance for series and parallel combinations.