Consider the boost converter circuit shown. In steady state, the inductor cur rent rises linearly from 0 A to 6 A in the first 10 µs and then falls linearly from 6 A to 0 A in the next 10 µs. The load resistance R is 10 Ω. What is the input voltage V_dc?

The correct option is 7.5 V.

1. Identify the circuit and waveform parameters from the image:
From the circuit diagram and the inductor current waveform in the image, we can extract the following values:
- Load resistance, $R=10\Omega$
- Peak inductor current, $I_{\text{peak}}=6\text{A}$
- Inductor current rise time (switch ON time), $T_{\text{on}}=10\mathrm{\mu s}$
- Inductor current fall time (switch OFF time), $T_{\text{off}}=10\mathrm{\mu s}$
- Total switching period, $T=T_{\text{on}}+T_{\text{off}}=10\mathrm{\mu s}+10\mathrm{\mu s}=20\mathrm{\mu s}$

2. Apply the inductor volt-second balance:
In steady-state operation, the average voltage across the inductor over one complete period is zero:

$$V_{\text{dc}}\cdot T_{\text{on}}+(V_{\text{dc}}-V_o)\cdot T_{\text{off}}=0$$
Substitute the values of $T_{\text{on}}=10\mathrm{\mu s}$ and $T_{\text{off}}=10\mathrm{\mu s}$:

$$V_{\text{dc}}\cdot 10+(V_{\text{dc}}-V_o)\cdot 10=0$$
$$2V_{\text{dc}}=V_o\Rightarrow V_o=2V_{\text{dc}}$$

3. Calculate the average load current:
In steady state, the average current through the output capacitor $C$ is zero. Therefore, the average load current $I_o$ must equal the average current through the diode $I_D$.
Since the diode only conducts during the off-period, the diode current waveform is a triangle with a peak value of $6\text{A}$ and duration $T_{\text{off}}=10\mathrm{\mu s}$:

$$I_o=I_{D,\text{avg}}=\frac{\text{Area under diode current waveform}}{T}$$
$$I_o=\frac{\frac{1}{2}\cdot I_{\text{peak}}\cdot T_{\text{off}}}{T}$$
Substitute the values:

$$I_o=\frac{\frac{1}{2}\cdot 6\cdot 10\cdot {10}^{-6}}{20\cdot {10}^{-6}}=\frac{30}{20}=1.5\text{A}$$

4. Find the average output voltage and input voltage:
Using Ohm's Law, the average output voltage across the resistor is:

$$V_o=I_o\cdot R=1.5\text{A}\cdot 10\Omega =15\text{V}$$
Using the relation $V_o=2V_{\text{dc}}$ from step 2, we solve for $V_{\text{dc}}$:

$$V_{\text{dc}}=\frac{V_o}{2}=\frac{15}{2}=7.5\text{V}$$