1.

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»

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



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