# Initial Conditions

Capacitor's initial condition need to be examined before charging a capacitor. Capacitor may have some charges stored, if it has charged before but have not fully discharged. In all the examples examined in the previous sections, the capacitor was uncharged before the switch was thrown. We will now examine the effect of a charge, and therefore a voltage (V = Q/C), on the plates at the instant the switching action takes place. Fig. 1: Defining the regions associated with a transient response.
The voltage across the capacitor at this instant is called the initial value, as shown for the general waveform of Fig. 1. Once the switch is thrown, the transient phase will commence until a leveling off occurs after five time constants. This region of relatively fixed value that follows the transient response is called the steady-state region, and the resulting value is called the steady-state or final value. The steady-state value is found by simply substituting the open-circuit equivalent for the capacitor and finding the voltage across the plates. Using the transient equation developed in the previous section, an equation for the voltage $v_C$ can be written for the entire time interval of Fig. 1; that is, $$v_C = V_i + (V_f - V_i)(1 - e^{-t/\tau})$$ However, by multiplying through and rearranging terms: $$v_C = V_i + V_f - V_i - V_f e^{-t/\tau}+V_i e^{-t/\tau}$$ $$\bbox[10px,border:1px solid grey]{v_C = V_f + (V_i -V_f)e^{-t/\tau}} \tag{1}$$ If you are required to draw the waveform for the voltage $v_C$ from the initial value to the final value, start by drawing a line at the initial and steady-state levels, and then add the transient response (sensitive to the time constant) between the two levels. The example to follow will clarify the procedure.
Example 1: The capacitor of Fig. 2 has an initial voltage of 4 V.
a. Find the mathematical expression for the voltage across the capacitor once the switch is closed.
b. Find the mathematical expression for the current during the transient period.
c. Sketch the waveform for each from initial value to final value. Fig. 2: Example 1
Solution:
a. Substituting the open-circuit equivalent for the capacitor will result in a final or steady-state voltage $v_C$ of 24 V. The time constant is determined by $$\begin{split} \tau &= (R_1 + R_2)C\\ &=(2.2 kΩ + 1.2 kΩ)(3.3 mF) = 11.22 ms \end{split}$$ with $$5 \tau = 56.1 ms$$ Applying Eq. (1): $$\begin{split} v_C &= V_f + (V_i - V_f)e ^{-t/\tau}\\ &= 24 V + (4 V - 24 V)e^{-t/11.22ms}\\ &= 24 V - 20 Ve^{-t/11.22ms} \end{split}$$ b. Since the voltage across the capacitor is constant at 4 V prior to the closing of the switch, the current (whose level is sensitive only to changes in voltage across the capacitor) must have an initial value of 0 mA. At the instant the switch is closed, the voltage across the capacitor cannot change instantaneously, so the voltage across the resistive elements at this instant is the applied voltage less the initial voltage across the capacitor. The resulting peak current is $$\begin{split} Im &= {E - v_C \over R_1 + R_2} \\ &= {24-4 \over 2.2kΩ+ 1.2kΩ} = 5.88 mA \end{split}$$ The current will then decay (with the same time constant as the voltage $v_C$) to zero because the capacitor is approaching its open-circuit equivalence. The equation for $i_C$ is therefore: $$iC = 5.88 mAe^{-t/11.22m}$$ c. Fig. 3: $v_C$ and $i_C$ for the network of Fig. 2