Answer:
[tex]V_{2.2} =2.2V[/tex]
Fast and loose: that's a classic voltage divider. the drop from the i-th resistor is [tex]R_i \over {\sum R}[/tex] of the drop across the whole series. In our situation, it's [tex]2.2\cdot 10^3 \over {2.2\cdot10^3 + 6.8\cdot 10^3[/tex] of 9 V. By plugging numbers in a calculator, it's 22/90 of 9V, or 2.2V
With Ohm's Law
The series is equivalent of a single resistor of Resistance [tex]2.2\cdot 10^3+6.8\cdot 10^3 \Omega = 9.0 k\Omega[/tex]. By Ohm's first Law ([tex]V=Ri[/tex]) the current flowing through the resistor is [tex]9V = 9*10^3\Omega i \rightarrow i=1mA[/tex]. At this point, the drop across the first resistor is, again by Ohm's law[tex]V = 2.2 \cdot 10^3 \Omega \cdot 1\cdot 10^{-3} A = 2.2V[/tex]
