Use the voltage-divider equation to find V_{\mathrm{out}} in Figure F–4. (V_{\mathrm{out}} is the voltage from the point labeled V_{\mathrm{out}} to the ground symbol.)
V_{\mathrm{out}}=12\,\mathrm{V}\times{\frac{2\,\mathrm{k}\Omega}{2\,\mathrm{k}\Omega\,+\,4\,\mathrm{k}\Omega}}
=4\,\mathrm{V}