P 7.8-8 The circuit shown in Figure P 7.8-8 is at steady state when the switch opens at time

t = 0.  Determine , , , , , , and , .

 

Figure P 7.8-8

Annotated Circuit from before switch is opened

At steady state the inductor acts like a short circuit and the capacitor acts like an open circuit.  So we know immediately that

 

 

 

So we have a simple series circuit which means

 

 

We can find

 

Now when the switch opens our circuit becomes

Since the left side of the circuit is open, we know

 

We also know from continuity

 

And

 

Now because current in inductor must be 4/3 A, we can find the voltage across the 6 Ω resistor

 

 

Now we can do a voltage walk to find the voltage across the inductor

 

 

 

Finally current through capacitor must be the same as the current in the inductor and the

6 Ω resistor because we have a simple series loop.  But since the current is going opposite to the drawn reference direction we have

 

 

So the answers are :

 

 

 

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This page last updated on January 12, 2019