Question

If certain number of bulbs rated at ($({P_1}\,{\text{watt,V}}\,{\text{volt),}}({P_2}\,{\text{watt,V}}\,{\text{volt)}}....$ are connected in series across a potential difference of V volt, then power ‘P’ consumed by all bulbs is given by,
A) $P = {P_1} + {P_2} + {P_3} +. ...... + {P_n}$.
B) $\dfrac{1}{P} = \dfrac{1}{{{P_1}}} + \dfrac{1}{{{P_2}}} + \dfrac{1}{{{P_3}}} +. .....$.
C) ${P^2} = {P^2}_1 + {P_2}^2 + {P_3}^2 +. ......$.
D) None of the above.

Answer 1

Hint
Relate the resistance of a bulb to its power by the formula $P = \dfrac{{{V^2}}}{R}$. Once done generalize it for n bulbs and find the equivalent power by using the equation of equivalent resistance i.e.${R_{eq}} = {R_1} + {R_2} +. ......$.

Complete step by step answer
Given that there are a certain number of bulbs which are connected in series. Let’s say n bulbs. Their power and voltage are also given as $({P_1}\,{\text{watt,V}}\,{\text{volt),}}({P_2}\,{\text{watt,V}}\,{\text{volt)}}....$
We know that for n bulbs the equivalent resistance in series is given by the formula,
${R_{eq}} = {R_1} + {R_2} +. ......$
We know that power is inversely related to resistance by the equation,
$P = \dfrac{{{V^2}}}{R}$
i.e. $R = \dfrac{{{V^2}}}{P}$
Now from above power expression, we have
$\Rightarrow {R_1} = \dfrac{{{V^2}}}{{{P_1}}}$
( Note that the voltage given is same for all the bulbs and also because the circuit is in series)
$\Rightarrow {R_2} = \dfrac{{{V^2}}}{{{P_2}}}$ and so on for ${R_3},{R_4}$etc.
On putting the values of the resistances in the equation ${R_{eq}} = {R_1} + {R_2} +. ......$ we have,
$\Rightarrow \dfrac{{{V^2}}}{{{P_{eq}}}} = \dfrac{{{V^2}}}{{{P_1}}} + \dfrac{{{V^2}}}{{{P_2}}} + \dfrac{{{V^2}}}{{{P_3}}} +. ........ = \dfrac{{{V^2}}}{{{P_n}}}$
Cancelling the term ${V^2}$ from the numerator we have,
$\Rightarrow \dfrac{1}{P} = \dfrac{1}{{{P_1}}} + \dfrac{1}{{{P_2}}} + \dfrac{1}{{{P_3}}} +. .....$
Hence, the power expression is given by $\dfrac{1}{P} = \dfrac{1}{{{P_1}}} + \dfrac{1}{{{P_2}}} + \dfrac{1}{{{P_3}}} +. .....$ and the option (B) is correct.

Note
If the question asked us the same question in parallel connection, it is easy to guess that the answer would be $P = {P_1} + {P_2} + {P_3} +. ...... + {P_n}$.
This is because resistance and power share reciprocal relationships. If one knows the expression for resistance, he can easily find the expression for power.