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:

$$P_T=P_1+P_2+P_3$$

Where $P_T$ is the total power.

Expert Answer

Given that:

$$R_1=400ohm$$

$$R_1=800ohm$$

Voltage is:

$$V=120V$$

We know that:

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

So, for the first bulb, we have:

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

By putting in the values, we get:

$$P_1=\frac{120^2}{400}$$

$$P_1=\frac{14400}{400}$$

$$P_1=36W$$

Now for the second bulb, we have:

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

By putting in the values, we get:

$$P_1=\frac{120^2}{800}$$

$$P_1=\frac{14400}{800}$$

$$P_1=18W$$

Numerical Answer

The power dissipated in the first bulb is:

$$P_1=36W$$

And for the second bulb, the power dissipated is:

$$P_1=18W$$

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:

$$R_1=600ohm$$

$$R_1=1200ohm$$

Voltage is:

$$V=120V$$

We know that:

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

So, for the first bulb, we have:

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

By putting in the values, we get:

$$P_1=\frac{120^2}{600}$$

$$P_1=\frac{14400}{600}$$

$$P_1=24W$$

Now for the second bulb, we have:

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

By putting in the values, we get:

$$P_1=\frac{120^2}{1200}$$

$$P_1=\frac{14400}{1200}$$

$$P_1=12W$$

Thus, the power dissipated in the first bulb is:

$$P_1=24W$$

And for the second bulb, the power dissipated is:

$$P_1=12W$$

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