Question

A bulb is rated 330V-110W. What do you think is its resistance? Three such bulbs burn for 5hrs at a stretch. What is the energy consumed? Calculate the cost in rupees if the rate is 70 paisa per unit?

Answer 1

We have been given the values of voltage and power and to find the resistance we know the formula which relates all the three terms which is power is equal to voltage squared divided by resistance. To find the energy consumed we know the formula, energy is equal to power into time. Finally, to find the cost we multiply the cost with the total energy consumed.

Complete answer:
We have been given in the question that,
$V=300V$
$P=110W$
We know that for any given electrical appliance, the power is given by the formula,
$Power=\dfrac{{{(Voltage)}^{2}}}{Resistance}$
$P=\dfrac{{{V}^{2}}}{R}$
Hence, the resistance will be,
$R=\dfrac{{{V}^{2}}}{P}$
Now substituting the values of P and V in the formula we get resistance as,
$R=\dfrac{{{(330)}^{2}}}{110}=\dfrac{108900}{110}$
$R=990\Omega$
Thus, the resistance of the bulb is $990\Omega$.
Now we will find the energy consumed by 3 bulbs burning at a stretch of 5 Hours,
We have been given that,
$P=110W$
$t=5hrs$
$n=3$bulbs
We know the formula for energy which is,
$Energy=n\times power\times time$
$E=nPt$
Now substituting the values in the formula, we get,

& E=3\times 110\times 5 \\\ & E=1650Wh \\\ \end{aligned}$$ Now we have to convert $$Wh$$ to $$kWh$$ which is the SI unit, $$\begin{aligned} & 1kWh=1000Wh \\\ & 1Wh=\dfrac{1}{1000}kWh \\\ & 1650Wh=\dfrac{1650}{1000}kWh=1.65kWh \\\ \end{aligned}$$ Therefore, the energy consumed by 3 bulbs is $$1.65kWh$$. Now let us look into the cost of $$1.65kWh$$ of energy, We have been given that cost per unit or 1$$kWh$$of energy is 70 paise, So, total cost for $$1.65kWh$$of energy =$$0.7\times 1.65=1.155$$ Therefore, the total cost of $$1.65kWh$$ of energy is Rs. 1.155. Hence, solved. **Note:** We have to read the question thoroughly as it has three sub-parts. While solving we have to keep in mind to keep all the units in the SI unit form. For any appliance, the power rating and energy consumption becomes the major factor.