Consider the buck-boost converter shown. Switch Q is operating at 25 kHz and 0.75 duty-cycle. Assume diode and switch to be ideal. Under steady-state condition, the average current flowing through the inductor is ______ A.

The correct answer is 24.

Based on the provided circuit diagram, we can identify the following components and parameter values:
- Supply voltage, $V_{\mathrm{dc}}=20V$
- Inductance, $L=1mH$
- Capacitance, $C=100\mu F$
- Load resistance, $R=10\Omega$
- Switching frequency, $f=25kHz$
- Duty cycle, $D=0.75$

Here is the circuit schematic of the buck-boost converter:

Step 1: Calculate the output voltage ($V_0$) of the buck-boost converter
For an ideal buck-boost converter operating under steady-state conditions, the relationship between the average output voltage magnitude and the supply voltage is given by:
$V_0=V_{\mathrm{dc}}·\frac{D}{1-D}$
Substituting the given values:
$V_0=20·\frac{0.75}{1-0.75}=20·\frac{0.75}{0.25}=20·3=60V$

Step 2: Calculate the average load current ($I_0$)
Using Ohm's law, the average current flowing through the load resistor is:
$I_0=\frac{V_0}{R}$
Substituting the calculated load voltage and resistance:
$I_0=\frac{60}{10}=6A$

Step 3: Calculate the average inductor current ($I_{L,\mathrm{avg}}$)
In a buck-boost converter, the inductor is connected to the load only during the off-state of the switch, which represents a fraction $(1-D)$ of the total switching period. Therefore, the relationship between the average load current and the average inductor current is:
$I_{L,\mathrm{avg}}=\frac{I_0}{1-D}$
Substituting the values:
$I_{L,\mathrm{avg}}=\frac{6}{1-0.75}=\frac{6}{0.25}=24A$