Question F.8: Calculate Vout in Figure F–8....

Because I = 0 A,

V_{\mathrm{drop(4k \Omega)}}=0\,\mathrm{A}\times4\,\mathrm{k}\Omega\,=\,0\,\mathrm{V}

V_{\mathrm{out}} =12\,\mathrm{V}\,-\,\mathrm{V_{drop}}\,=\,12\,\mathrm{V}

Note:
This is probably the hardest concept to understand. Another way to explain why the entire supply reaches V_{\mathrm{out}} is to assume that an open circuit can be modeled by an extremely large resistance, let’s say, 10 MΩ. If you then apply the voltage-divider equation to the circuit with 10 MΩ in place of the open, the calculation will be

V_{\mathrm{out}}=\;12\,\mathrm{V}\times{\frac{10,002,000}{10,002,000+4,000}}=\;11.995\,\mathrm{V}