Electronic – Getting started with Power Dissipation Problem

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'.

A rather convoluted question, but I think what you need is an adjustable voltage source with a minimum output impedance of 5k.

In that case, why don't you design your voltage divider to give you the correct output voltage and then buffer it like so: