waveform of load current iL  if the switch is closed for t=0 to t=100µS and then opened. Consider all the circuit components ideal and inductor value is 200µH.

1. Break Down the Problem

We need to analyze the waveform of the load current $i_L$ in an electrical circuit when a switch is closed for a specified time and then opened. Given:

• Inductor value $L = 200 \, \mu H$
• Switch closed from $t = 0$ to $t = 100 \, \mu s$

2. Relevant Concepts

When the switch is closed, the inductor will start charging and the current through it will increase exponentially. The current through an inductor is given by:

$i_L(t) = i_0 \left(1 - e^{-\frac{t}{\tau}}\right)$

where $\tau = \frac{L}{R}$ is the time constant of the RL circuit, $i_0$ is the final steady-state current, and $R$ is the resistance in the circuit. We will need to determine the value of $R$ or consider it as an ideal case.

3. Analysis and Detail

1. Current Increasing Phase (Switch Closed):

   • Since no resistance $R$ is mentioned, we will assume there is some resistance in the circuit, or else the current won't saturate. Let’s assume a steady state occurs at: $i_L(100 \, \mu s) = i_0 \left(1 - e^{-\frac{100 \, \mu s}{\tau}}\right)$ If we let $i_0$ be the maximum steady state current that will eventually be reached.

2. Current Decreasing Phase (Switch Opened):

   • Once the switch is opened, the inductor will maintain the current flowing for a brief period until all energy is dissipated: $i_L(t) = i_{\text{final}} e^{-\frac{t - 100 \, \mu s}{\tau}}$ where $i_{\text{final}}$ is the value of $i_L$ right before the switch is opened.

4. Verify and Summarize

Without specific values for the load resistance, we can't compute specific current values, but we can summarize the behavior:

• For $t = 0$ to $t = 100 \, \mu s$, $i_L$ increases toward a steady state defined by $R$.
• After $100 \, \mu s$, $i_L$ will start to decrease exponentially.

Final Answer

• The load current $i_L(t)$ during $0 \leq t < 100 \, \mu s$ can be modeled as: $i_L(t) = i_0 \left(1 - e^{-\frac{t}{\tau}}\right)$
• After $t = 100 \, \mu s$: $i_L(t) = i_{\text{final}} e^{-\frac{t - 100 \, \mu s}{\tau}}$