Consider the boost converter shown. Switch Q is operating at 25 kHz with a duty cycle of 0.6. Assume the diode and switch to be ideal. Under steady-state condition, the average resistance R_in as seen by the source is ________ Ω . (Round off to 2 decimal places)

To find the average resistance $R_{\mathrm{in}}$ as seen by the source, we analyze the boost converter under steady-state conditions in continuous conduction mode (CCM).

From the given schematic, we identify the following parameter values:
- Source voltage, $V_s=15\text{V}$
- Duty cycle, $D=0.6$
- Load resistance, $R=10\text{Ω}$
- Inductance, $L=1\text{mH}$
- Capacitance, $C=100\text{μF}$
- Switching frequency, $f=25\text{kHz}$

For an ideal boost converter operating in steady-state CCM, the relation between the output voltage ($V_o$) and the input source voltage ($V_s$) is given by:
$V_o=\frac{V_s}{1-D}$

Assuming an ideal (lossless) converter, the average input power supplied by the source must equal the average output power delivered to the load:
$P_{\mathrm{in}}=P_{\mathrm{out}}$
$\frac{V_s^2}{R_{\mathrm{in}}}=\frac{V_o^2}{R}$

By substituting the relationship for $V_o$ into the power equation:
$\frac{V_s^2}{R_{\mathrm{in}}}=\frac{{\left(\frac{V_s}{1-D}\right)}^2}{R}$
$\frac{V_s^2}{R_{\mathrm{in}}}=\frac{V_s^2}{R{\left(1-D\right)}^2}$

Simplifying the equation gives the relationship for the average input resistance:
$R_{\mathrm{in}}=R{\left(1-D\right)}^2$

Now, we substitute the given values of load resistance $R=10\text{Ω}$ and duty cycle $D=0.6$ into the formula:
$R_{\mathrm{in}}=10\times {\left(1-0.6\right)}^2$
$R_{\mathrm{in}}=10\times {\left(0.4\right)}^2$
$R_{\mathrm{in}}=10\times 0.16=1.6\text{Ω}$

Thus, the average resistance $R_{\mathrm{in}}$ as seen by the source is 1.6 Ω.