Question

Two bulbs one of 200 volts, 60 watts the other of 200 volts, 100 watts are connected in series to a \[200\]-volt supply. The power consumed will be:
A 37.5 watt
B 160 watt
C 62.5 watt
D 110 watt

Answer 1

Hint: Power consumption refers to the electrical energy that is supplied per unit time, such as a home appliance, to operate something. In units of watts, power consumption is usually measured. In calculating your energy consumption, the first step is to figure out how many watts are used per day by each device. Just multiply the wattage of your appliance by the number of hours that you use it in a day. This will give you the number of watt-hours each day consumed.

Formula used:
$\text{P}=\text{VI}$

Complete step by step solution:
Here, two bulbs one of 200 volts, 60 watts $\&$ the other of 200 volts, 100 watts are connected in series to a 200-volt supply.
For bulb 1,
$\mathrm{I}=\dfrac{\mathrm{V}}{\mathrm{P}}=\dfrac{60}{200}=\dfrac{6}{20} \mathrm{~A}$
Hence,
$\mathrm{R}=\dfrac{\mathrm{V}}{\mathrm{I}}=\dfrac{200 \times 20}{6}=\dfrac{4000}{6} \Omega$
Similarly, for bulb 2 ,
$\mathrm{I}=\dfrac{\mathrm{V}}{\mathrm{P}}=\dfrac{100}{200}=\dfrac{1}{2} \mathrm{~A}$
Hence,
$\mathrm{R}=\dfrac{\mathrm{V}}{\mathrm{I}}=\dfrac{200 \times 2}{1}=400 \Omega$
When bulbs are connected in series their resistances will added up
$\Rightarrow {{\text{R}}^{\prime }}=\left( \dfrac{4000}{6} \right)+400$
$\mathrm{R}^{\prime}=\dfrac{6400}{6}$
Hence, current is
$\mathrm{I}^{\prime}=\dfrac{\mathrm{V}}{\mathrm{R}^{\prime}}=\dfrac{200 \times 6}{6400}=\dfrac{12}{64} \mathrm{~A}$
Hence, $\mathrm{P}^{\prime}=\mathrm{V} \mathrm{I}=200 \times \dfrac{12}{64}$
\[\therefore \]$\mathrm{P}^{\prime}=37.5 \mathrm{W}$
The power consumed will be $\mathrm{P}^{\prime}=37.5 \mathrm{W}$

The correct option is (a).

Note:
The same amount of current flows through each of them since the bulbs are in series. Note now that the potential in this case is not 200 volts across each bulb. Instead, throughout the combination of bulbs, it is the voltage. Thus, the power they consume significantly decreases when the bulbs are connected in series. The voltage rating is also enhanced by a series link. There is a voltage limit of 200 or 300 volts for tiny wattage resistors, regardless of power dissipation. So, if you need a higher voltage, connect several in series across the resistor.