Heaters in parallel and series I have a heater rated at 13 kW with two identical heating elements connected in parallel with a 240VAC power supply. If I connect elements in series instead what should be my expectation:

*

*That the total resistance in the circuit will increase thereby reducing the current for the same power dissipation?

or


*The power dissipation will increase due to the increased total resistance?

 A: You are correct that the resistance will increase.
If you take the resistance of one element to be $R$, then in parallel in the effective resistance is $\frac{1}{2}R$, and ins series it is $2R$, so quantitatively, it increases by a factor of four.
Now, does this increased resistance mean that more power will be dissipated? Well this depends on your type of power supply.
If you have a constant current power supply, then the voltage drop over the resistor is $IR$, and $I$ does not depend on $V$ (it is fixed by the power supply). Therefore the power dissipated across the resistor $IV_{resistor}$ is $I^2R$. Therefore the power output increases by a factor of four for a constant current power supply.
On the other hand, if you have a constant voltage power supply, then the current through the circuit is dependent on the resistance. It is $I = \frac{V_{source}}{R}$. From here we can follow the same arguments as before to get the power output as $I^2R$. But since $I$ depends on $R$ now, we have that the power output for a constant voltage power supply is $(\frac{V_{source}}{R})^2R = \frac{V_{source}^2}{R}$. Therefore the power output $decreases$ by a factor of four for a constant voltage power supply.
In summary, the power output of the heater is dependent entirely on the current flowing through it and its resistance. Once you know the current you can calculate the potential difference across the heater using the resistance and combine this with the original current flowing through it to get the power output. Most heaters will use a constant voltage power supply, so the current is determined by the circuit's resistance. In this case, that is the heater and we find that the power output is inversely proportional to the heater's total resistance, $\frac{V_{source}^2}{R}$.
