P 8.3-2 The circuit shown in Figure P 8.3-2 is at steady state before the switch opens at time

t = 0. The input to the circuit is the voltage of the voltage source, 12 V. The output of this circuit is the current in the inductor, i(t). Determine i(t) for t > 0.

 

Answer: i(t) = 1 + e–0.5t A for t > 0

 

P8_3_2

Figure P 8.3-2

 

The General Solution is of the Form:

 

 

Where x is the variable we are looking for usually voltage for a capacitor  and current for an inductor.  The “∞” means the value of the variable after a long time, for voltage, this is usually referred to as the “Open Circuit” voltage.  For current, this is usually referred to as the “Short Circuit” current.  Finally τ is the appropriate time constant.  For a capacitor

 

For an inductor

So for an inductor, The solution looks like:

 

 

Before the switch opens the circuit looks like

In steady state, the inductor is acting like a short-circuit so that the current is found simply from

 

After the switch has been opened for a while the circuit looks like

Again the right resistor is shorted out in time so the long term current is

 

 

Deactivating the power source we have a circuit that looks like

So we have two resistors in series yielding

 

This is now in parallel so our Thevenin resistance is

 

 

So

 

Putting these values into our standard form yields

 

 

Simplifying

 

 

 

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This page last updated on June 28, 2019