The symbol for a battery is seldom drawn in schematic diagrams. Figure F–9 is an alternative schematic for a series circuit. Solve for V_{\mathrm{out}}.
V_{\mathrm{out}}=12\,\mathrm{V}\times{\frac{2\,\mathrm{k}\Omega}{2\,\mathrm{k}\Omega\,+\,4\,\mathrm{k}\Omega}}
= 4 V