As Seekonk points out, I'm not sure if this is what you're trying to do.
In the case of the first circuit, initially, there is no charge across the cap so the voltage across the "resistor" (the relay coil), will be your battery voltage. Therefore current will flow, and the relay will close. This current will begin charging up the capacitor and so the voltage across capacitor will increase, the voltage across the "resistor" will drop and eventually the current will become too low to turn on the relay, and the relay will turn off.
In the case of the second circuit, you can pretty much ignore the capacitor because it is in parallel with the "resistor" and so the voltage across the "resistor" is always your battery voltage so current will always flow and your relay will always be closed. The only time the capacitor is interesting is when you remove the battery after some time. In this case, the capacitor will contain enough charge to keep current flowing in the "resistor" for a while, so the relay will stay closed for some time before opening..