A Question on Resistance
Your math is correct. Since we have 

Ohm's Law \$V = IR\$

For the lightbulb: 

\$12V = 100mA * R\$

Solving for R we get 

\$R = \frac{12V}{0.1A}\$
or
\$R = 120\Omega\$

Adding a \$70\Omega\$ resistor in series makes the total resistance across the 19V supply \$190\Omega\$. 

Using \$V = IR\$ again, we can see that 
\$I = \frac{V}{R}\$ so the current flowing through the series circuit is \$\frac{19V}{190\Omega}\ = 0.1A \$ as shown by the ammeter. 

We can also double check the voltage across the lightbulb is 12V. 

I would guess the 100mA is a typical or MAX value, and by increasing the series resistor (R1) you could dim the lightbulb, possibly extending it's life. 







simulate this circuit &ndash; Schematic created using CircuitLab