# Power consumption in simple example

power

Let's say I have 100W light bulb connected to battery. So let's say the lightbulb will use 100W power.

Now let's say I have 2 light bulbs:

First 100W

Second 200W

What is the power use of each light bulb if they are connetcted in series circuit? I mean cause it is series circuit will they get enough power or just part of what they "need"?

Suppose that the declared power is a nominal voltage of 12V, and the series circuit is powered by a 12 V battery.
In this case, each lamp can be modeled with a resistor:

$$R_{L1} = \dfrac{12^2}{100} = 1.44\,\Omega$$

$$R_{L2} = \dfrac{12^2}{200} = 720\,\mathrm{m}\Omega$$

That is, when we apply to each lamp 12 V, the power consumption is the nominal (100 W and 200 W), and this power corresponds to the calculated resistance values​​.

For simplicity, we assume that the resistance is linear (although in reality depends on factors such as the temperature of the filament). Connecting these two resistors in series, we obtain a total resistance:

$$R_{eq} = R_{L1} + R_{L2} = 2.16\,\Omega$$

so that the circuit current flows as

$$I = \dfrac{12}{R_{eq}} = 5.556 \,\mathrm{A}$$

This current, dissipates power in each resistor as:

$$P_{L1} = I^2\cdot R_{L1} = 44.44\,\mathrm{W}\\ P_{L2} = I^2\cdot R_{L2} = 22.22\,\mathrm{W}$$

As you can see, on each lamp, less than the rated power is developed, which results in lower brightness. It is important to remember that these results are only approximate because a lamp is not a completely linear element.

As additional information, we can obtain the necessary current to each lamp at rated:

$$I_{L1nom} = \dfrac{100}{12} = 8.333\,\mathrm{A}\\ I_{L2nom} = \dfrac{200}{12} = 16.67\,\mathrm{A}\\$$

which are much higher than those estimated for the series connection of two lamps values.

Remember, this is only one example at a given supply voltage, but can serve as a guide for your analysis.