Suppose that in the circuit shown in <image 1>, $V_{CC}$ = 24 V, RB = 960k$\Omega$, RC = 7 k$\Omega$, $\beta $ = 80, and the leakage is egligible. What is the value of of $V_{CE}$, where $V_{CE}$ is the collector-to-emitter voltage ?
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With negligible ICBO, we first find that, with VBE small,

                                    IB = (VCC / RB) = (24V / 960k\Omega) = 25\muA

                        We now solve for the collector

                                    IC = \betaIB = 80(25\muA) = 2mA

                        By kirchhoff's low VCC = VCE + ICRC; therefore

                                    VCE = VCC - ICRC

                        and in this case

                                    VCE = 24 V - (2 mA) (7k\Omega) = 24 - 14 = 10 V