Resistor in parallel of battery pack

I have a 25,000mah solar rechargeable battery pack powering a Nano. It outputs 5v steadily.

I want to power the nano and a small vibrating dc motor that I will have vibrate 30 seconds every 6 hours. So basically it will vibrate for 2 minutes every day.

I need to add a resistor I'm parallel to the battery pack to keep the pack from "shutting down" after a few seconds of low current draw.

My question is, if I have a 33 ohm 1 watt resistors, how do I calculate how much energy will be lost to the battery pack every hour or so?

I didn't post the schematic or code because I don't have it yet. I want to calculate power loss first to see if it's feasible.

So 5v/33ohms = 0.151 A * 5v = 0.755Watts

So at rest 3/4 of a watt should be dissipated just fine by the 1w R.

Right?

Well, I doubt you need 5V/33 ohm = 151mA to keep your battery pack alive.
Most discussed here need much less, and then only a shot of current every once in a while, which you can have the Arduino sleep & wake up to turn a transistor on briefly as needed.

How much current does your motor use when on?
Do you need the full Nano, or could you use a Promini and just attach an FDTI Basic for programming it to save more current?

Marciokoko:
I have a 25,000mah solar rechargeable battery pack powering a Nano. It outputs 5v steadily.

I want to power the nano and a small vibrating dc motor that I will have vibrate 30 seconds every 6 hours. So basically it will vibrate for 2 minutes every day.

I need to add a resistor I'm parallel to the battery pack to keep the pack from "shutting down" after a few seconds of low current draw.

My question is, if I have a 33 ohm 1 watt resistors, how do I calculate how much energy will be lost to the battery pack every hour or so?

I didn't post the schematic or code because I don't have it yet. I want to calculate power loss first to see if it's feasible.

So 5v/33ohms = 0.151 A * 5v = 0.755Watts

So at rest 3/4 of a watt should be dissipated just fine by the 1w R.

Right?

The way to keep a power bank like this from shutting down is to draw a current spike regularly, every 20 seconds
or whatever it takes to prevent it cutting out - pull enough continuous current and you'll flatten the thing all too quickly.

And I very much doubt you actually have 25Ah pack, that's smacks of Chinese amps. Perhaps its a cheap one that really 2.2Ah?

Please provide links for all of your hardware so we know what exactly you have...

I also believe there are ways to defeat the power-down feature too, but that is invasive and
might be tricky.

Well I want the nano because I plan to expand depending on my power availability in real terms (once I determine it).

It's a Chinese bought :

https://goo.gl/images/TRNbF9

That’s a 15V pack, not 5V

Well but the reading with a multimeter says 5v. I guess that's a general link to their products and then you can select the voltage.

Or would it be simpler to bypass the circuitry of the battery pack responsible for the sleep part?

So I have this setup (on the left) and instead of using the resistor in parallel to the motor load, im thinking of a circuit to keep the battery pack "awake". So I found this link in a post here in Arduino.cc:

BatteryPackLoad

But Im not sure how to fit the circuit into my existing setup. Ive added the circuit suggested to the right here. Is that switch supposed to be the pin from my nano?

Bueller, Bueller...

Can someone explain how this oscillator circuit works?