Hi,

I am thinking of building an electric rat trap.

The basic components will be: 12v battery, uf 200 capacitor.

MY QUESTION IS :
When the circuit is closed and the energy is released from the fully charged capacitor (charged to 12v from the battery) is the current likely to kill rodents.

The resistance on the circuit is about 0.75 ohms (measured using a test meter).

Ohms law suggests that there should be a big current being discharged by the capacitor, so in theory this should zap the rat!

I've seen some designs on YT where people have used capacitors charged to 500v etc, but this seems to be a bit excessive to me.

Surely this 12v circuit should deliver a current big enough to get the job done, shouldn't it?

It this scenario, what is the capacitor doing. The rapid discharge of energy must make this device more potent than using just a 12v battery. Is this correct thinking?

My knowledge of electronic is only basic. What's your thoughts on this please?

thanks
Zaffy

 

You need a lot more than 12V to zap a rat.
Note that you can touch both12V battery terminals and not feel anything.
A body (of you or the rat) has too much resistance for 12V to generate a significant current through it.
Typically you have to get above about 50V before you can even feel the current.

 

The current is 12V divided by the resistance of the rat.
To get an idea what the resistance of a rat might be, measure the resistance with your test meter from your thumb to your little finger.

 

I have dismantled a couple of failed high voltage rat killer devices. The voltage is always at least 1000 volts. The reason is that the connection to the rat is very poor, and also the rat may flinch at the first hint of a shock. In addition, the high voltage must deliver enough current over a long enough time to fatally damage the rat.
So what is needed is a narrow place that the rat must squeeze through, touching both electrodes, and the a trigger beyond that space so the shock will pass through vital parts of the rat.

 

Not to mention that the energy of a 200uF capacitor charged to 12V is 0.0144 joules. Compare that to the same 200uF cap charged to 1000V, which is 200 joules. At 12V it is probably not enough to kill the flea on the rat’s back.

Bob

 

My knowledge of electronic is only basic. What's your thoughts on this please?

12V isn't a lethal voltage for the type of contact you envision.

 

The big deal part is that to do enough damage takes a lot more power than just enough to sometimes do some damage if you are lucky. That is why a serious rat trap will almost mash a rat in half. Keep in mind that the rat will not cooperate, and so things must be way overdone.
My "Rat-a-pault"rat destroyer sent them backing out of the barrel with a 100PSI blast of compressed air, at which time the rat meets some carefully placed sharp edges. That system, intended for grain warehouse use, provides both self-cleaning and way overkill as means to assure destruction. But it should be able to kill 50 or more rats every hour.

I created the design in response to a plea for some means to keep the rats from eating so much grain in the warehoses where it was arriving to feed starving people. The benefit is that it automaticly cleans itself and resets, and mostly does not consume much power during standby operation. So one 12 volt battery could power many of them for a week or two. It does require compressed air, but not at a high flow rate, so that a single compressor can power a whole warehouse with 50 of these devices. The only required labor is removing the pile of dead rats, which is not a skilled labor sort of job.
Unfortunately the whole concept is so very different from what very small minds can deal with that the UN people would not even consider it.

 

A 12 volt supply with a billion amps available will - - - a wet rat with a body resistance of just 100 ohms will draw ( 12V ÷ 100Ω = ) 0.012 amps (12mA {milli-amps}). 12mA isn't going to accomplish anything. 120V under the same circumstances will produce 120mA. It's iffy at best. 1200 volts under the same circumstances will produce 1.2 amps. That's fatal for anything that touches it. Wet of course. IF the rat is dry and dusty his body resistance may be 1,000 to 10,000Ω. You've seen the math. You should be able to figure it out from there.

 

Thanks for all those who contributed to this thread.
I think I probably didn't grasp the true implications of resistance, which makes this a flawed design.
I will look at how I can step up 12v to 500+v

 

very small minds can deal with that the UN people would not even consider it.

Yes, unfortunately there are people out there who see them as cute little furry guys. In my case they have caused devastation in outbuildings so I need to eradicate them

 

It's iffy at best. 1200 volts under the same circumstances will produce 1.2 amps.

Yes, the resistance problem. Without any real resistance the 12v would get the job done.But I suppose the world doesn't always work in our favour!

 

Not to mention that the energy of a 200uF capacitor charged to 12V is 0.0144 joules. Compare that to the same 200uF cap charged to 1000V, which is 200 joules. At 12V it is probably not enough to kill the flea on the rat’s back.

Bob

Yes, true. They'd probably enjoy the little tingle!

 

The current is 12V divided by the resistance of the rat.
To get an idea what the resistance of a rat might be, measure the resistance with your test meter from your thumb to your little finger.

I forgot that when we were kids we used to put our tongues across the terminals of 9v batteries!

 

12V isn't a lethal voltage for the type of contact you envision.

My revised design will be lethal I'm sure!

 

I will look at how I can step up 12v to 500+v

Note that due to the conservation of energy, you’ll need a supply capable of much higher current. Just for an example, let’s say you stepped 12V up to 960V… to get 1A of current at 960V, you’d need a 12V supply that could supply 80A.

The law is that power in must equal or be greater than the power out. Ignoring losses due to the transformation, 12x80 = 960W in. That will result in 960x1 = 960W out.

But there is losses in stepping the voltage up. So you might only get 0.8A out.

 

My revised design will be lethal I'm sure!

When you approach that point, you'll likely be in violation of site rules:

 

Yes, true. They'd probably enjoy the little tingle!

You are welcome to produce a "Ratapault"for your own use, as many as it takes to solve the problem. They can be powered by AC mains or 24 volts or even a 12 volt battery. They also need compressed air to pressurize the tank that launches the rat into some sharp objects. No electronics, just a bit of wiring switches and a solenoid valve.

 

The only required labor is removing the pile of dead rats, which is not a skilled labor sort of job.

I wonder who got that job.