P 8.4-5 The circuit shown in Figure P 8.4-5 is at steady state before the switch opens at t = 0. The switch remains open for 0.5 second and then closes. Determine v(t) for t ≥ 0.

 

P8_4_5

Figure P 8.4-5

 

If the circuit is in steady state by t = 0, then the capacitor is fully charged and acting like an open circuit and we have the following circuit

The voltage across the capacitor is the same as the voltage across the parallel pair of resistors from the 10 Ω and the 40 Ω resistors on the right of the capacitor.  So combining them the effective resistance is

 

 

So we have an effective resistance of 8 Ω.  The voltage across that which is the same as the voltage across the capacitor can be determined by voltage division since the 8 Ω resistor is in series with the remaining 40 Ω resistor and the power supply.

 

 

When the switch opens if the circuit reached steady state, it would look like.

So the voltage for the open circuit of the capacitor this time is just the voltage across the right 40 Ω resistor.  Since the circuit is now just two identical resistors in series, the voltage division give us

 

The Thevenin resistance is found from deactivating the power source

So the two resistors are in parallel, so

 

 

So the standard form for the capacitor voltage is

 

 

 

Now when the switch closes again at t = 0.50 s, we need to know what the capacitor voltage was so plug into that equation and we get

 

 

This becomes our new “v(0)” value.  The long term steady state value, we already know from above is 4 V.  We do need a new Thevenin resistance which now comes from the circuit

This time the Thevenin resistance is due to all three resistors being in parallel, so

 

So

Again using the standard equation

 

 

 

 

 

Please send any comments or questions about this page to ddonovan@nmu.edu

This page last updated on July 11, 2019