Answer

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**Hint:**In order to solve this question, the knowledge of the calculation of required resistance that is the resultant resistance of circuit is important. This should be kept in mind that the formula for calculation of resultant resistance is different for parallel and series combination.

**Complete step by step answer:**

It is given in question that,

Power of ${1^{st}}$ lamp, ${P_1} = 40W$ and voltage of ${1^{st}}$ lamp, ${V_1} = 220V$

Power of ${2^{nd}}$ lamp, ${P_2} = 60W$ , voltage of ${2^{nd}}$ lamp , ${V_2} = 220V$

As we know that,

$P = \dfrac{{{V^2}}}{R}$

Where $P$ is the power, $V$ is the potential difference and $R$ is the resistance.

In terms of resistance we have,

$R = \dfrac{{{V^2}}}{P}$

So, the resistance of ${1^{st}}$ lamp, would be given as,

${R_1} = \dfrac{{V_1^2}}{{{P_1}}}$

Putting the values of the respective quantities in the above equation we have,

${R_1} = \dfrac{{{{(220)}^2}}}{{60}}$

On Solving we have,

${R_1} = \dfrac{{2420}}{3}\Omega $

Now the resistance of the second lamp would be calculated the same way and would be given as,

${R_2} = \dfrac{{V_2^2}}{{{P_2}}}$

Putting the values of the respective quantities in the above equation we have,

${R_2} = \dfrac{{{{(220)}^2}}}{{40}}$

On Solving we have,

${R_1} = 1210\Omega $

(A) In the a part of the question both the lamps are in parallel connection,

So, the required resistance would be given by,

$\dfrac{1}{{\operatorname{Re} q}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$

Putting the respective values we have,

$\dfrac{1}{{\operatorname{Re} q}} = \dfrac{3}{{2420}} + \dfrac{1}{{1210}}$

On simplifying the above expression we get,

$\dfrac{1}{{\operatorname{Re} q}} = \dfrac{5}{{2420}}$

On solving we get,

${R_{req}} = 484\Omega $

As we know, the current drawn from electrical supply is given as,

$I = \dfrac{V}{{{R_{req}}}}$

Putting the values we have,

$I = \dfrac{{220}}{{484}}$

On solving we get,

$I = \dfrac{5}{{11}}A$

(B) Energy consumed by the two lamps in one hour is equal to energy consumed by ${1^{st}}$ lamp in one hour + energy consumed by ${2^{nd}}$ lamp in one hour

The equation would be represented as,

$E = {P_1} \times 1 + {P_2} \times 1$

This can also be written as,

$E = ({P_1} + {P_2}) \times 1$

Putting the values of ${P_1}$ and ${P_2}$ we have,

$E = (60W + 40W) \times 1hour$

On solving we have,

$E = 100Wh$

As, energy is expressed in $KWh$ and $1KWh = {10^2}Wh$

We have,

$E = 0.1KWh$

**Note:**The required resistance in series connection is calculated by simply adding the values of all resistance the given resistors in the series combination while in parallel connection as given in the question the reciprocal of the required resistance is equal to the sum of reciprocals of the resistances.

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