Modeling with First Order DE’s — Question 5

Question 5 – Solution

Strategy. Use the circuit balance backward: a specified capacitor trajectory determines its current and therefore the required source.

Step 1: Derive the voltage model. Since $I=CU'$, the loop equation gives $$RC\,U'+U=E(t),\qquad U(0)=0.$$ Here $RC=(1000)(0.001)=1$ s is the time constant. With time numerically measured in seconds, the model is $U'+U=E$; the derivative coefficient carries the one-second time constant.

Step 2: Find the required input and current. The desired derivative is $U_d'=3e^{-t/2}$ V/s, so $$\boxed{E(t)=6-3e^{-t/2}\text{ V}},\qquad \boxed{I(t)=0.003e^{-t/2}\text{ A}}.$$ Substitution gives $RC\,U_d'+U_d=E$ and $U_d(0)=0$. The prescribed $U_d$ fixes its derivative, so the loop relation leaves no other possible source trajectory. In particular, the initial current is $3$ mA and the required initial source is $3$ V.

Step 3: Apply the source constraint. The required source increases strictly from $3$ V toward $6$ V. It first reaches $5$ V when $e^{-t/2}=1/3$. Therefore the full feasible initial interval is $$\boxed{0\le t\le t_*=2\ln 3\approx 2.197\text{ s}},\qquad \boxed{U_d(t_*)=4\text{ V}}.$$ Equality at the limiting time is allowed by the given source bound. Beyond it the unique required source exceeds $5$ V.

Step 4: Compute energy and interpret saturation. At the limiting time, $$\boxed{W=\tfrac 12(0.001)(4)^2=0.008\text{ J}}.$$ A source fixed at $5$ V afterward would instead produce $U=5-e^{-(t-t_*)}$ V, starting from $4$ V. It tends to $5$ V, whereas the desired trajectory tends to $6$ V. Thus the circuit can continue operating, but it cannot continue this exact target under the source constraint. The feasible interval is a hardware limitation, not a singularity of the ideal differential equation.