Question 19.SP.8: A RC series circuit with a rectangular pulse input is shown ......
A RC series circuit with a rectangular pulse input is shown (Fig. 19-13). A rectangular pulse is a special case of a constant dc source because the voltage is a constant V when the pulse is on and a constant zero when the pulse is off. If a single pulse is applied to the circuit, find the voltage across the resistance at the time of 0.5, 1, and 2 ms. The pulse has a maximum voltage of 10 V and lasts for 1 ms.
Find the time constant of the circuit.
T=R C=\left(1 \times 10^3\right)\left(1 \times 10^{-6}\right)=1 \times 10^{-3}=1\ msWrite the formula for v_{R} when the circuit is charging.
v_R=V e^{-t / R C} (19-16)
When t = 0.5 ms,
v_R=10 e^{-0.5 / 1}=10 e^{-0.5}=10(0.607)=6.07\ VWhen t = 1 ms,
v_R=10 e^{-1 / 1}=10 e^{-1}=10(0.368)=3.68\ VAt t = 1 ms, the pulse is turned off. At that instant the source voltage is zero, so
\begin{aligned}v_R+v_C & =0 \\ v_R & =-v_C\end{aligned}Since the voltage across the capacitor v_C cannot change instantly, the voltage across the resistor becomes -v_C. Now at an instant before t = 1 ms,
\begin{aligned}v_R+v_C & =10 \\ v_C & =10-v_R=10-3.68=6.32\ V\end{aligned}Therefore, at t = 1 ms, v_R=-v_{C}=-6.32 V and then v_R decays to zero. So the formula for v_R at t = 1 ms is:
v_R=-6.32 e^{-t / R C}Then in the next 1 ms or when t = 2 ms measured from the origin,
v_R=-6.32 e^{-1}=-6.32(0.368)=-2.33\ VThe plot for v_R is as shown in Fig. 19-14.
