Question

Two bulbs rated 50w-230V and 100W-230V are given. Which one of the bulb will be brighter when they are connected to the main in series first and in parallel later?

Answer 1

Given:

Rating of the 1st bulb = 50 W - 230 V

Rating of the 2nd bulb = 100 W - 230 V

To Find :

Which bulb will glow brighter when they are connected at first in series and then in parallel  :

Solution :

We know that :

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

Where, P denotes  power , V  voltage and R  resistance of the bulbs .

So, for the 1st bulb :

$P_1 = 50W$

Or, $50W = \frac{(230)^2}{R_1}$

Or, $R_1 = \frac{(230)^2}{50W}$

Similarly for 2nd bulb:

$R_2=\frac{(230)^2}{100W}$

On diving $R_1$ and $R_2$ :

$\frac{R_1}{R_2}= \frac{\frac{(230)^2}{50}}{\frac{(230)^2}{100}}$

Or, $\frac{R_1}{R_2}= 2$

So, $R_1=2R_2$

For series combination:

By Ohm's Law the ratio of voltage drop :

$\frac{V_1}{V_2}= \frac{R_1}{R_2} = 2$

As, Power =$\frac{V^2}{R}$

So, $\frac{P_1}{P_2}= \frac{\frac{V_1^2}{R_1}}{\frac{V_2^2}{R_2}}=(\frac{V_1}{V_2})^2 \times \frac{R_2}{R_1}$

           =$2^2\times \frac{1}{2} = 2:1$

Here in this case $P_1$ is double of $P_2$ b , so $P_1$ is greater , hence 50 W bulb will glow brighter .

Now for the parallel connection :

Voltage across the two bulbs in parallel connection will be same , so :

$\frac{V_1}{V_2} = 1$

So, $\frac{P_1}{P_2}= \frac{\frac{V_1^2}{R_1}}{\frac{V_2^2}{R_2}}= \frac{V_1}{V_2}\times \frac{R_2}{R_1} = 1 :2$

Hence, $\frac{P_1}{P_2}=\frac{1}{2}$

So , in this case $P_2$ is greater , hence 100 W bulb will glow brighter  .

Hence, in series connection 50 W bulb glows brighter and in parallel connection 100 W bulb glows brighter