Question

Two exactly similar electric lamps are arranged (i) parallel and (ii) in series. If the parallel and series combination of lamps are connected to $220\,{\text{V}}$ supply line one by one, what will be the ratio of electric power consumed by them?

Answer 1

The question has two cases, in the first case the electric lamps are connected in parallel and in the second case the electric lamps are connected in series. Find the equivalent resistance and power consumed in each case. To find the equivalent resistance you will need to recall the formula for equivalent resistance in parallel and in series and use these formulas to find the equivalent resistance in each case. Then find the power consumption in each case and find their ratio.

Formulas used:
For two resistances in parallel, we calculate the equivalent resistance by the formula,
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ (i)
where ${R_1}$ and ${R_2}$ are the two resistances.
For two resistances in series, we calculate the equivalent resistance by the formula,
${R_{eq}} = {R_1} + {R_2}$ (ii)
where ${R_1}$ and ${R_2}$ are the two resistances.
The power consumption can be written as,
$P = \dfrac{{{V^2}}}{{{R_{eq}}}}$ (iii)
where $V$ is the voltage supplied and ${R_{eq}}$ is the equivalent resistance.

Complete step by step answer:
Given, supply voltage, $V = 220\,{\text{V}}$.And the two electric lamps are similar.As the electric lamps are exactly similar their resistance will be the same. Let the resistance of each of the electric lamps be $R$. Here, ${R_1} = R$ and ${R_2} = R$,Electric lamps in parallel.
Here, we use the formula from equation (i) and put the values of ${R_1}$ and ${R_2}$ to get the equivalent resistance.
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{R} + \dfrac{1}{R}$
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{2}{R}$
$\Rightarrow {R_{eq}} = \dfrac{R}{2}$
The power consumption in this case will be (using equation (iii))
$P = \dfrac{{{V^2}}}{{{R_{eq}}}}$
Putting the values of $V$ and ${R_{eq}}$ we get,
$P = \dfrac{{{{220}^2}}}{{\left( {\dfrac{R}{2}} \right)}}$
$\Rightarrow P = \dfrac{{96800}}{R}$ (iv)
Electric lamps are in series
Here, we use the formula from equation (ii) and put the values of ${R_1}$ and ${R_2}$ to get the equivalent resistance.
${R_{eq}}^\prime = R + R$
$\Rightarrow {R_{eq}}^\prime = 2R$
The power consumption in this case using equation (iii) we get,
$P' = \dfrac{{{V^2}}}{{{R_{eq}}^\prime }}$
Putting the values of $V$ and ${R_{eq}}^\prime$ we get,
$P' = \dfrac{{{{220}^2}}}{{2R}}$
$\Rightarrow P' = \dfrac{{24200}}{R}$ (v)
Now, dividing equation (iv) by (v) we get,
$\dfrac{P}{{P'}} = \dfrac{{\left( {\dfrac{{96800}}{R}} \right)}}{{\left( {\dfrac{{24200}}{R}} \right)}}$
$\Rightarrow \dfrac{P}{{P'}} = \dfrac{{96800}}{{24200}}$
$\Rightarrow \dfrac{P}{{P'}} = \dfrac{4}{1}$
$\therefore P:P' = 4:1$

Therefore, the ratio between the electric powers consumed by electric lamps in parallel to electric lamps in series is $4:1$.

Note: Most of the time students get confused between the formulas for equivalent resistance in parallel and in series so, carefully remember the two formulas. Another point to remember is power consumption in a circuit increases if the resistance in the circuit is reduced, that is power consumption is inversely proportional to the resistance in the circuit.