Question

Two bulbs of 250 V and 100 W are first connected in series and then in parallel with a supply of 250 V. Total power in each of the case will be respectively
A.) 100W, 50W
B.) 50W, 100W
C.) 200W, 150W
D.) 50W, 200W

Answer 1

Hint: In order to find the power in both the circuit we need to first find the equivalent resistance in between the two circuits. When there are two circuits in which one is in parallel connection and the other is in series connection then the difference between them is of equivalent resistance and power depends on the voltage and resistance.

Formula used:
$P=\dfrac{{{V}^{2}}}{R}$
${{R}_{eq}}={{R}_{1}}+{{R}_{2}}$(in series)
$\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$(In parallel)

Complete Step by Step Solution:
In series connection.
We know that the equivalent resistance in series circuit is calculated from formula ${{R}_{eq}}={{R}_{1}}+{{R}_{2}}$, but for that we need each of the resistance in the circuit.

Here we are given only with potential difference(V) and power(P). so here we will use the formula $P=\dfrac{{{V}^{2}}}{R}$
Where P = Power of bulb
R = Resistance of the circuit
V = Potential difference
$\Rightarrow {{P}_{1}}=\dfrac{{{V}^{2}}}{{{R}_{1}}}$
$\Rightarrow {{R}_{1}}=\dfrac{{{V}^{2}}}{{{P}_{1}}}$
$\Rightarrow {{R}_{1}}=\dfrac{{{(250)}^{2}}}{100}$
$\Rightarrow {{R}_{1}}=625\Omega $

Similarly, we will calculate resistance for second bulb
$\Rightarrow {{R}_{2}}=\dfrac{{{V}^{2}}}{{{P}_{2}}}$
$\Rightarrow {{R}_{2}}=\dfrac{{{(250)}^{2}}}{100}$
$\Rightarrow {{R}_{2}}=625\Omega $

Equivalent resistance will be ${{R}_{eq}}={{R}_{1}}+{{R}_{2}}$
$\Rightarrow {{R}_{eq}}=625+625$
$\Rightarrow {{R}_{eq}}=1250\Omega $

Power for series circuit
$\Rightarrow {{P}_{s}}=\dfrac{{{V}^{2}}}{{{R}_{eq}}}$
$\Rightarrow {{P}_{s}}=\dfrac{{{(250)}^{2}}}{1250}$
$\Rightarrow {{P}_{s}}=50W$

In parallel circuit
$\Rightarrow {{P}_{1}}=\dfrac{{{V}^{2}}}{{{R}_{1}}}$
${{R}_{1}}=\dfrac{{{V}^{2}}}{{{P}_{1}}}$
$\Rightarrow {{R}_{1}}=\dfrac{{{(250)}^{2}}}{100}$
$\Rightarrow {{R}_{1}}=625\Omega $

And for second bulb
${{P}_{2}}=\dfrac{{{V}^{2}}}{{{R}_{2}}}$
$\Rightarrow {{R}_{2}}=\dfrac{{{V}^{2}}}{{{P}_{2}}}$
$\Rightarrow {{R}_{2}}=\dfrac{{{(250)}^{2}}}{100}$

We have $\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$
$\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{625}+\dfrac{1}{625}$
$\Rightarrow {{R}_{eq}}=312.5\Omega $

Power for parallel circuit
${{P}_{p}}=\dfrac{{{V}^{2}}}{{{R}_{eq}}}$
$\Rightarrow {{P}_{p}}=\dfrac{{{(250)}^{2}}}{312.5}$
$\Rightarrow {{P}_{p}}=200W$
From above calculations we got ${{P}_{s}}=50W$and ${{P}_{p}}=200W$

So, we can conclude that option (D) is the correct answer.

Note:
Since we do not have resistance for each of the bulbs or each of the circuits is given and hence, we need to use the power formula to calculate the resistance first and then calculate the equivalent resistance to get the total resistance for each type of circuit. After getting equivalent resistance we were able to find the power for each type of circuit