# Two light bulbs have constant resistances of 400 ohm and 800 ohm. If the two light bulbs are connected in series across a 120 V line, find the power dissipated in each bulb

The main objective of this question is to find the power dissipated in each bulb that is connected in series.

This question uses the concept of power in series. In a series circuit, the total power is the same as the total amount of power lost by each resistor. Mathematically, it is represented as:

$\space P_T \space = \space P_1 \space + \space P_2 \space + \space P_3$

Where $P_T$ is the total power.

Given that:

$\space R_1 \space = \space 400 \space ohm$

$\space R_1 \space = \space 800 \space ohm$

Voltage is:

$\space V \space = \space 1 2 0 \space V$

We know that:

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

So, for the first bulb, we have:

$\space P_1 \space = \space \frac{V^2}{R_1}$

By putting in the values, we get:

$\space P_1 \space = \space \frac{1 2 0^2}{4 0 0}$

$\space P_1 \space = \space \frac{1 4 4 0 0}{4 0 0}$

$\space P_1 \space = \space 3 6 \space W$

Now for the second bulb, we have:

$\space P_2 \space = \space \frac{V^2}{R_2}$

By putting in the values, we get:

$\space P_1 \space = \space \frac{1 2 0^2}{8 0 0}$

$\space P_1 \space = \space \frac{1 4 4 0 0}{8 0 0}$

$\space P_1 \space = \space 1 8 \space W$

The power dissipated in the first bulb is:

$\space P_1 \space = \space 3 6 \space W$

And for the second bulb, the power dissipated is:

$\space P_1 \space = \space 1 8 \space W$

## Example

In the above question, if the resistance across one bulb is $600$  ohm and 1200  ohm across another bulb. Find the power dissipated along these two bulbs which are connected in series.

Given that:

$\space R_1 \space = \space 6 0 0 \space ohm$

$\space R_1 \space = \space 1 2 0 0 \space ohm$

Voltage is:

$\space V \space = \space 1 2 0 \space V$

We know that:

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

So, for the first bulb, we have:

$\space P_1 \space = \space \frac{V^2}{R_1}$

By putting in the values, we get:

$\space P_1 \space = \space \frac{1 2 0^2}{6 0 0}$

$\space P_1 \space = \space \frac{1 4 4 0 0}{6 0 0}$

$\space P_1 \space = \space 24 \space W$

Now for the second bulb, we have:

$\space P_2 \space = \space \frac{V^2}{R_2}$

By putting in the values, we get:

$\space P_1 \space = \space \frac{1 2 0^2}{1 2 0 0}$

$\space P_1 \space = \space \frac{1 4 4 0 0}{1 2 0 0}$

$\space P_1 \space = \space 1 2 \space W$

Thus, the power dissipated in the first bulb is:

$\space P_1 \space = \space 2 4 \space W$

And for the second bulb, the power dissipated is:

$\space P_1 \space = \space 1 2 \space W$