P 8.3-1 The circuit shown in Figure P 8.3-1 is at steady state before the switch closes 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 voltage across the capacitor, v(t). Determine v(t) for t > 0.

 

Answer: v(t) = 6 – 2e–1.33t V for t > 0

 

P8_3_1

Figure P 8.3-1

 

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 a capacitor, The solution looks like:

 

 

For  the circuit looks like

Assuming the circuit is in steady state, the voltage across the capacitor is the voltage across the right 6 Ω resistor.  Since there are 3 6 Ω resistors in series with the capacitor acting like an open circuit, voltage division gives us

 

 

After the switch is closed our circuit becomes

This time we only have two 6 Ω resistors in series, so doing voltage division again we find that the long term voltage across the capacitor is

 

 

Deactivating power source so we can calculate the Thevenin Resistance we get a circuit

So the Thevenin resistance is two 6 Ω resistors in parallel and

 

 

So

 

Putting these values into our standard form yields

 

simplifying

 

 

 

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This page last updated on June 28, 2019