Do I need a resistor if \$\frac{V}{I}=0\$? 

I am working on a simple circuit of an LED and a buzzer in series where the voltage drop would be 2 and 3 volts, so \$R=\frac{V}{I}=\frac{(5 - (3 + 2))}{I}\$, in that case do I need a resistance?

Would a current flow in the previous case? Isn't current flow due to the voltage difference of maybe 5V and 0V on a 5V battery? So when I drop the voltage of 5V, why would current even flow?

Also in the following picture:

![enter image description here][1]


  [1]: https://i.sstatic.net/SETNj.jpg


In this picture, that 98.90A is where current tends to infinity as resistance is too low, but in the first one, how did it calculate that 49.22mA ?

\$R = \frac{(5-2)}{49mA} = 61 ohms\$, how did Circuit Wizard add 61 ohms, isn't that way too high for wire resistance?