Question

Find the energy in kWh consumed in 10 hours by four devices of power 500W each
a) 30 kWh
b) 25 kWh
c) 20 kWh
d) 35 kWh

Answer 1

Hint : Let us say the above devices are connected as the way they are connected in our house. Let the devices draw their own current hence, all the four devices be connected in parallel. Hence we can find the net energy consumed from power dissipated in a parallel circuit.

Complete solution:
Let us first understand how the power dissipated across resistors in series and parallel vary
First let us understand for series
Let two resistors if resistance R be connected in series and connected to a battery of emf E as shown in the fig

The resistance in series gets added up hence net resistance is equal to ${{R}^{'}}={{R}_{1}}+{{R}_{2}}$
The power dissipated in an electrical circuit is given by,
$P=\dfrac{{{V}^{2}}}{R}$ …..(1)
Let us denote the power dissipated in the individual resistors as, ${{P}_{1}}and{{P}_{2}}$
Substituting the equivalent resistance in equation 1,
$P=\dfrac{{{V}^{2}}}{{{R}_{1}}+{{R}_{2}}}$
Cross multiplying the above equation we get,
$\dfrac{{{R}_{1}}+{{R}_{2}}}{{{V}^{2}}}=\dfrac{1}{P}$
Splitting ${{V}^{2}}$in the denominator the above equation becomes,
$\dfrac{{{R}_{1}}}{{{V}^{2}}}+\dfrac{{{R}_{2}}}{{{V}^{2}}}=\dfrac{1}{P}$
Taking R1 and R2 in the denominator,
$\dfrac{1}{\dfrac{{{V}^{2}}}{{{R}_{1}}}}+\dfrac{1}{\dfrac{{{V}^{2}}}{{{R}_{2}}}}=\dfrac{1}{P}$
From the definition of power let us write the power dissipated across individual resistances in the above equation
$\dfrac{1}{{{P}_{1}}}+\dfrac{1}{{{P}_{2}}}=\dfrac{1}{P}$
If n resistances are connected in parallel than the power dissipated across them is,
$\dfrac{1}{P}=\dfrac{1}{{{P}_{1}}}+\dfrac{1}{{{P}_{2}}}.......\dfrac{1}{{{P}_{n}}}$
Similarly let us derive an expression for power dissipated across in parallel

The resistance in parallel resistance is equal to $\dfrac{1}{{{R}^{'}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$
Let us denote the power dissipated in the individual resistors as, ${{P}_{1}}and{{P}_{2}}$
Substituting the equivalent resistance in equation 1,

$P={{V}^{2}}\left( \dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} \right)$
Multiplying ${{V}^{2}}$in the numerator, the above equation becomes,
$P=\dfrac{{{V}^{2}}}{{{R}_{1}}}+\dfrac{{{V}^{2}}}{{{R}_{2}}}$

From the definition of power let us write the power dissipated across individual resistances in the above equation
$P={{P}_{1}}+{{P}_{2}}$
If n resistances are connected in parallel than the power dissipated across them is,
$P={{P}_{1}}+{{P}_{2}}........{{P}_{n}}$
Now since we have equation for power dissipated across resistance in parallel, the power dissipated across the four resistances is 4 time the power of each i.e.
P= 4(500)
P=2000 W
Energy consumed in an electrical circuit is given by
$E=P\times t$ where P is the power dissipated and t is the time for which the power is consumed.
Therefore the energy consumed by the four devices is $E=P\times t=20\times {{10}^{3}}=20kWh$

Hence, the correct answer to the above question is c.

Note: We have assumed the devices are connected in parallel as they are connected in our houses. But if the devices were connected in series the power dissipated across them would have been different. Power dissipated across n equal resistances in series equals $P=\dfrac{p}{n}$ and in parallel equals $P=pn$ where p is the power dissipated across each of the resistors.