Question 10.SP.13: A lossless buck-boost converter with continuous inductor cur......

(a) Solve (10.9) for D to find
G_{V} = {\frac{V_{2}}{V_{1}}} = {\frac{D}{1  –  D}}               (10.9)

(b) Based on (1.23) for this lossless converter,
P_{\mathrm{dc}} = {\frac{V_{\mathrm{dc}}^{2}}{R}}                (1.23)
P_{in} = P_{o} = {\frac{V_{2}^{2}}{R_{L}}} = {\frac{(15)^{2}}{10}} = 22.5\,\mathrm{W}

(c) The average value of input current follows as
I_{1} = {\frac{P_{in}}{V_{1}}} = {\frac{22.5}{12}} = 1.875\,\mathrm{A}