Power consumption in simple example
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.