Outline: 
The Transient Response of RC Circuits 
The Transient Response of RL Circuits 
The Complete Response 
Analysis Steps for finding the Complete Response of RC and RL Circuits 
Study Problems 
In this circuit, there is a pulse, a resistor, and a capacitor. Assume here that the pulse goes from 10V down to 0V at t=0. Assume also that the circuit is in Steady State at t=0. This implies that the capacitor is 'open' at t=0. In order for KVL to be true at t=0 then the capacitor voltage must be 10V at t=0. This is because there is no current in the circuit, therefore the voltage across the resistor is zero. V_{c}(0) = V_{c}(0+) = 10V 

Note that since the Transient Response is the circuit's response to energies stored in storage elements, we will 'kill' the pulse source. This leaves us with a simple ResitorCapacitor circuit with an initial 10V on the capacitor at t=0+. Applying KCL to an RC circuit: Cdv/dt + V/R = 0 dv/dt + V/(RC) = 0 ∫dv/V = ∫1/(RC) dt ln V = t/(RC) + K ln V(t=0) = K ln V_{o} = K ←V_{o} is the voltage on the cap at t=0+. lnV  ln V_{o} = t/(RC) ln (V/V_{o}) = t/(RC) V/V_{o} = e^{t/(RC)} V(t) = V_{o} e^{t/(RC)} ←V_{o} = 10V in this example. Note that the speed at which the capacitor discharges from 10V to 0V is determined by the product R×C When t=RC, the voltage on the capacitor is V_{o}/e or 37% of it's initial value. We call RC the time constant and the symbol is τ For an RC circuit, τ=RC In this particular circuit τ = RC = 100Ω×1mF = 0.1 seconds This means it takes 0.1 seconds for the capacitor to discharge from 10V down to 3.7V. Looking at the response at the right, does this look about right? 

Here is the same circuit as that above, except that the resistor value is doubled. This means that τ is also doubled. τ = RC = 200Ω×1mF = 0.2 seconds This circuit is twice as slow as the last circuit. 

Compare this response to the last one. Does it appear that it takes twice as long for the capacitor to discharge? 

Finding the Time Constant τ: Using an oscilloscope or PSpice, we can calculate τ by inspecting the response curve. Locate the place where V = 37% × V_{o}. Then find the time at which this occurs. This time is the time constant, τ. 
In this circuit, there is a pulse, a resistor, and an inductor. Assume here that the pulse goes from 10V to 0V at t=0. Assume also that the circuit is in Steady State at t=0. This implies that the inductor is a 'short' at t=0. In order for KCL to be true at t=0 the inductor current must be 1A at t=0. I_{L}(0) = I_{L}(0+) = 1A Consider the circuit at t=0+, the voltage across the pulse is zero but since I_{L}(0+) = 1A then V_{R} = 10V. Therefore for KVL to be true V_{L} = +10V. Therefore V_{L} = +10V is the initial voltage across the inductor. 

Note that since the Transient Response is the circuit's response to energies stored in storage elements, we will 'kill' the pulse source. This leaves us with a simple ResitorInductor circuit with an initial 10A going through the inductor at t=0+. Applying KVL to an RL circuit: iR + Ldi/dt = 0 iR/L + di/dt = 0 iR/L = di/dt R/L dt = di/i ∫R/L dt = ∫di/i Rt/L + K = ln i K = ln i(t=0) K = ln i_{o} Rt/L = ln i  K Rt/L = ln i  ln i_{o} Rt/L = ln(i/i_{o}) i/i_{o} = e^{Rt/L} i(t) = i_{o}e^{Rt/L} ←i_{o} in this case is 1A Since the plot on the right is for voltage we will find V_{L} using V_{L} = Ldi/dt V_{L} = (1H) d[i_{o}e^{Rt/L}]/dt = (1H) (10) i_{o}e^{Rt/L} V_{L} = 10 e^{Rt/L} When t=L/R, the voltage on the inductor is V_{o}/e or 37% of it's initial value. We call L/R the time constant and again the symbol is τ For an RL circuit, τ=L/R In this particular circuit τ = L/R = 1H/10Ω = 0.1 seconds This means it takes 0.1 seconds for the inductor to go from 10V down to 3.7V. Looking at the response at the right, does this look about right? 

Here is the same circuit as that above, except that the resistor value is halved. This means that τ is doubled. τ = L/R = 1H/5Ω = 0.2 seconds This circuit is twice as slow as the last circuit. 

Compare this response to the last one. Does it appear that it takes twice as long for the circuit to dissipate it's energy? 

Finding the Time Constant τ: Using an oscilloscope or PSpice, we can calculate τ by inspecting the response curve. Locate the place where V = 37% × V_{o}. Then find the time at which this occurs. This time is the time constant, τ. 
Here is an RC Circuit with a Forcing Function: Assume the source is a pulse which goes from 0V to 10V at t=0. If we assume steady state at t=0, then there is no initial energy stored in the circuit. Intuitively we know that the capacitor is going to charge up to 10V. When the capacitor gets to 10V then the circuit is again at steady state. The pulse is forcing the capacitor to 10V, thus the 10V on the capacitor is called the forced response. The time it takes the capacitor to charge up to 10V is determined by the time constant. The response of getting to 10V is the transient response. 

Now we will find the Complete Response for V across the capacitor. This equation will match the curve shown at the right. From the last section we know that the transient response for an RC circuit is: V(t) = V_{o} e^{t/(RC)} = A e^{t/(RC)} Note that A is just some constant. We also know from inspection that eventually the capacitor will charge up to 10V. Now putting the transient and forced responses together we get: V_{complete} = A e^{t/(RC)} + V_{forced} V_{complete} = A e^{t/(RC)} + 10V Now we need to find A such that the equation equals V_{o} at t=0. In other words, the equation must satisfy the initial condition. V(t=0) = 0, therefore: 0 = A e^{0} + 10V = A + 10V A = 10V V_{complete} = 10e^{t/(RC)} + 10V V_{complete} = 10e^{10t} + 10V Note that when t>>0, V_{complete} = 10V. This intuitively means that when the transient response is gone the forced response still remains. 

In this circuit, the capacitor DOES NOT start at 0V. In other words the capacitor has a nonzero initial condition of 5V: Note that the left switch closes at the same time the right switch opens. Intuitively we can see that the capacitor is going to start at 5V and then charge up to 15V. For t<0 the 5V source is the forcing function and for t>0 the 15V source is the forcing function. Since this is an RC circuit with a forcing function, the response takes the following form: V_{complete} = A e^{t/(RC)} + V_{forced} By inspection we know that V_{forced} = 15V V_{complete} = A e^{t/(RC)} + 15V Now we need to find A such that the entire equation satisfies the value of V at t=0. V(t=0) = 5V = A e^{0} + 15V A = 10V V_{complete} = 10e^{t/(RC)} + 15 V V_{complete} = 10e^{10t} + 15 V 

Here is the Complete Response: Does this curve match the equation: V_{complete} = 10e^{10t} + 15 V 

Now let's find the voltage across the resistor for the RL circuit to the right. Note that the pulse goes from 5V to 15V at t=0. Assume that the circuit is in steady state at t=0. At steady state inductors look like 'shorts' therefore the voltage across the resistor must be equal to the pulse voltage of 5V at t=0. At t=0+ the voltage across the resistor is still 5V, but WHY????? Since the current in the inductor is continuous from 0 to 0+, and the current in the resistor is the same as the current in the inductor, and the voltage across the resistor is determined by its current, then we can say that if the resistor's current is continuous then the resistor's voltage must also be continuous. At t>>0 the voltage across the resistor is 15V. For t>0 we eventually reach steady state (as the transient response dies away), so we know that at t>>0 the inductor will look like a 'short'. Therefore the voltage across the resistor will equal the voltage of the pulse. Therefore we have both the initial condition and the forced response for the voltage across the resistor: V_{o} = 5V V_{forced} = 15V 

Now we will find the Complete Response: V_{complete} = A e^{t/τ} + V_{forced} V_{complete} = A e^{Rt/L} + V_{forced} V_{complete} = A e^{Rt/L} + 15V To find A we must let t=0 and assign the equation to V_{o}: V(t=0) = 5V = A e^{0} + 15V = A + 15 A = 10 V_{complete} = 10 e^{Rt/L} + 15V V_{complete} = 10 e^{Rt/L} + 15V R/L = 5 V_{complete} = 10 e^{5t} + 15V 

Assume the pulse oscillates between 15V and 5V every 1 s. Consider the waveforms to the right. One is the pulse and the other is the voltage across the inductor. Note that as the transient response for V_{L} dies away, the voltage across the inductor falls to zero. This is because the inductor acts like a 'short' in steady state conditions. However immediately after the switch moves a voltage is induced across the inductor. This voltage opposes the change that is occuring to the current in the circuit. Apply KVL to the circuit at t=0+ and t=1+ and see if you can get the 10V and 10V initial conditions for V_{L}. 