Electronic – arduino – Understanding short circuits (in Arduino) and identifying electronic devices

The first two calculations look good. The first one is a little ambiguous on the problem's behalf because the current is more than 3A, but not a short circuit so we don't explicitly know the series resistance of the battery. I also agree with your answer for the last one but let's understand why.

Let's do the calculation exactly the same though for c. We now have a zero ohm resistor. And since we know a short circuit condition is present we know the equivalent series resistor will be 2Ω.

$$ I = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + \frac{1}{0}}\Omega + 2.0\Omega \right)} $$ Simplifies to: $$ I = \frac{12V}{\left( \frac{1}{\infty}\Omega + 2.0\Omega \right)} $$ Finally we can see the the equation does simplify to $$ I = \frac{12V}{2.0\Omega} = 6A $$

Appending to answer your last question which I missed. Even if you don't explicitly know what the purpose of the infinity resistor is doing in the circuit solve it the same way: $$ I = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + \frac{1}{\infty}}\Omega + 2.0\Omega \right)} = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + 0}\Omega + 2.0\Omega \right)} $$ Which is what you had.

**Side note, current is usually denoted with an 'I' not an 'A'.

It seems the problem is the sound is right but not loud enough. That means you aren't driving a big enough speaker hard enough.

You don't say what kind of speaker SPKR1 is, but you probably want a 4 Ω type meant for cars. The low impedance allows more power to be dumped into it from the relatively low 12 V.

That also means you have to beef up Q1. A 2N2222 is totally inappropriate here. I would get rid of R1 and replace Q1 with a FET like the IRLML2502. That would allow some decent current thru the speaker.

I can't even guess what you think R2 is doing for you, but it's not doing anything useful. Replace it with a resistor of about the same impedance as the speaker (around 4 Ω if you follow my speaker recommendation) and a Schottky diode in series. The diode must be oriented in "reverse" so that it does not conduct when the transistor is on.

The speaker acts in part like a inductor, so whatever current is flowing thru it right before the switch turns off will still flow thru it right after the switch turns off. This inductance will make whatever voltage is necessary to keep the current flowing in the short term. In your circuit, that includes abusing and eventually blowing out the transistor. The resistor and diode I describe above gives that current a nice and safe place to go. The resistor being about the same as the speaker impedance means the current will wind down when turned off about with about the same time lag is it winds up when turned on.