Huagong Circuit Principle Test Questions (preferably detailed with process)

Solution: When t<0, the circuit is in a steady state, the capacitor is equivalent to an open circuit, Uc(0-)=Us1=12V.

Changing path theorem: Uc () = Uc (0-) = 12V. The capacitance at the moment of switching is equivalent to a 12V voltage source. At this time: ic()×R+Uc()+Us2=0, ic()=(12+8)/3k=-20/3(mA).

When t=∞, the capacitor is equivalent to an open circuit again, ic(∞)=0, Uc(∞)=-Us2=-8V.

Looking from both ends of the capacitor, the equivalent resistance of the circuit is R=3kΩ, and the time constant τ=RC=3×1000×10/1000000=0.03 (s).

Sanya quick release: f(t)=f(∞)+[f()-f(∞)]e^(-t/τ).

ic(t)=(-20/3-0)e^(-t/0.03)=-20/3e^(-100t/3)? (A).

Uc(t)=-8+(12+8)e^(-t/0.03)=-8+20e^(-100t/3)? (V).