In the dc-dc converter circuit shown, switch Q is switched at a frequency of 10 kHz with a duty ratio of 0.6. All components of circuit are ideal and the initial current in the inductor is zero. Energy stored in the inductor in mJ (rounded off to 2 decimal places) at the end of 10 complete switching cycles is ______.

The correct answer is 5.

1. Understanding the Circuit Parameters from the Given Image:

From the provided circuit diagram, we can identify the following components and parameters:
- Input DC Voltage ($V_s$): 50 V
- Inductance ($L$): 10 mH = 10 × 10^-3 H
- Diode $D$ connected in series with another 50 V source across the inductor branch.
- Switching frequency ($f$): 10 kHz
- Duty ratio ($D$): 0.6
- Initial inductor current ($i\left(0\right)$): 0 A

2. Time Period Calculations:

The total switching time period ($T$) is given by:

$T=\frac{1}{f}=\frac{1}{10\times {10}^3}={10}^{-4}\text{ s}=0.1\text{ ms}$

The ON-time of switch Q ($T_{\text{on}}$) during each cycle is:

$T_{\text{on}}=D\times T=0.6\times 0.1\text{ ms}=0.06\text{ ms}=60\mu \text{s}$

The OFF-time of switch Q ($T_{\text{off}}$) during each cycle is:

$T_{\text{off}}=(1-D)\times T=0.4\times 0.1\text{ ms}=0.04\text{ ms}=40\mu \text{s}$

3. Operation Analysis per Cycle:

Mode 1: Switch Q is ON ($0\le t\le T_{\text{on}}$)
When switch Q is closed, the left 50 V source is connected directly across the 10 mH inductor. The diode $D$ is reverse biased by the secondary source.
The voltage across the inductor is $V_L=+50\text{ V}$.
The increase in inductor current during $T_{\text{on}}$ is:

$\Delta I_{\text{on}}=\frac{V_L}{L}\times T_{\text{on}}=\frac{50}{10\times {10}^{-3}}\times (60\times {10}^{-6})=5000\times 60\times {10}^{-6}=0.3\text{ A}$

Mode 2: Switch Q is OFF ($T_{\text{on}}<t\le T$)
When switch Q turns off, the inductor current freewheels through diode $D$ and the right-hand 50 V source.
The voltage across the inductor reverses such that $V_L=-50\text{ V}$.
The decrease in inductor current during $T_{\text{off}}$ is:

$\Delta I_{\text{off}}=\frac{50}{10\times {10}^{-3}}\times (40\times {10}^{-6})=0.2\text{ A}$

4. Net Current Change per Cycle:

The net change in current per switching cycle is:

$\Delta I_{\text{net}}=\Delta I_{\text{on}}-\Delta I_{\text{off}}=0.3\text{ A}-0.2\text{ A}=0.1\text{ A}$

5. Current after 10 Complete Switching Cycles:

Since the initial current is $i\left(0\right)=0\text{ A}$, the current at the end of 10 complete switching cycles will be:

$I\left(10\text{ cycles}\right)=10\times \Delta I_{\text{net}}=10\times 0.1\text{ A}=1.0\text{ A}$

6. Energy Stored in the Inductor:

The energy stored in an inductor is given by the formula:

$E=\frac{1}{2}L{I}^2$

Substituting the values of $L=10\text{ mH}$ and $I=1.0\text{ A}$:

$E=\frac{1}{2}\times (10\times {10}^{-3}\text{ H})\times (1.0\text{ A}{)}^2=5\times {10}^{-3}\text{ J}=5\text{ mJ}$

Thus, the energy stored in the inductor at the end of 10 complete switching cycles is 5 mJ.