You can connect the grounds in one point or leave them separate. As long as the only common point between the two circuits is the isolated power supply you are free to do as you wish.
If OTOH you want to connect them somewhere (to use the car antenna or sth.) then you should carefully choose the connection point (or even points) to avoid ground loops and ground noise.
You did the math right but overlooked one key issue, which is that all batteries have some internal resistance by themselves. As a rough first approximation, you can think of a battery as a ideal voltage source with some resistance in series. The short circuit current of the battery is therefore its voltage divided by that internal resistance.
9V batteries are particularly bad at internal resistance. You have a nearly dead one, which will be even worse. As a battery is used up, its voltage goes down and its internal resistance goes up.
Since this battery is pretty much dead anyway, you can have some fun with it without much loss. Let's characterize it according to the simplified model of it being a voltage source with series resistance, otherwise known as a "Thevenin source". First, measure the battery voltage without any load. Even a crappy voltmeter will have at least 1 MΩ resistance, which draws so little current that there is no effective voltage drop accross the internal resistor. Measuring the battery with a voltmeter therefore tells you its voltage source value directly.
To measure its internal resistance, you use two measurements of its voltage at different currents. If we assume for now that the internal voltage source puts out the same in both cases, the difference in external voltage is the difference in current times the internal resistance. You already have one measurement, which is essentially at 0 current. You could put a known resistance accross the battery to draw a few mA and see how much its external voltage drops as a result. Try it. In the extreme case you can short the battery by putting a ammeter accross it. At that point the current will be the internal voltage divided by the internal resistance. That's right in theory, but in reality the internal voltage source will drop due to some chemical effects at high current. It will also drop rapidly over time, so if you do this, you have to take the measurement quickly.
Other things you could do is to put a fixed resistance accross the battery and record the voltage over time as it goes completely dead. From that you can calculate and plot power output, total energy output, or just see the flatness or lack thereof of the voltage. Or measure the voltage shortly after connecting different resistances accross the battery. That lets you calculate internal resistance as a function of current. See how constant or not it is.
Batteries are not simple. Getting some intuition about them can be valuable.
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Old car horns are just big buzzers .An electromagnet is supposed to pull in a armature and open its contact cutting its power .when the open contact occurs the magnetic field dies away and the armature falls back to its rest position allowing for the cycle to keep repeating itself.When a horn is running the duty cycle of the current pulses is generally less than 50% .This means that the peak current is several times the average current.If you place too much resistance in the circuit you will find that the horn wont start but it can draw more than its rated current. I guess it would be possible to burn out a car horn this way.Most cars have a horn relay that is close to the battery .The horn switch can be further away.I havent tried this but i think a solid state switch would work.