Eventually the capacitor voltage will increase until

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + – iC = 0 + vC = VS – after a sufficiently long time 5. Eventually, the capacitor voltage will increase until it is equal to VS. At that point, the current drops to zero. From that time nothing changes again. Now, it doesn’t matter if the switch is closed or open. This is a transient effect. Initially, the capacitor voltage was at some fixed, steady value and the capacitor current was zero. When the switch closed, the current jumped up to a value determined by the difference between the source voltage, the initial capacitor voltage and the resistance connecting them. The current charged the capacitor to increasingly higher voltage. When the capacitor voltage reached the source voltage, the current dropped to zero, and the circuit to a static (quiescent) state with the capacitor at a new, higher voltage with no current flowing. EE 442 RC and RL transients – 5 R VS + – iR iC(t) C+ vC(t) – Now, let’s do the math. For t > 0 (after the switch has closed): KCL iC = iR V S − v C ( t) iR = R dvC (t) iC = C dt dvC (t) V S − v C ( t) C = dt R This is a differential equation. Fortunately, it is probably the first one you learned to do in your first diff. eq. class. EE 442 RC and RL transients – 6 dvC VS − v C C = dt R dvC dt = VS − v C RC Solve it by integrating both sides. ￿ vC (t) VCi ￿ dvC VS − ￿ vC = ￿ t o dt￿ RC The right side is time, and we integrate from t = 0 to t. (As usual, use a dummy variable in the integrand.) The left side is voltage, and the integration limits must match those on the time side. Therefore, we integrate from vC(t = 0) = VCi to vC(t). EE 442 RC and RL transients – 7 Carrying out the integration: ￿vC (t ) ￿ ￿ ￿t ￿ t￿ ￿￿ − ln (VS − vC ) ￿ = RC ￿o VCi and manipulating the result t − ln [VS − vC (t)] + ln [VS − VCi ] = RC ￿ ￿ VS − VCi t − ln = V S − v C ( t) RC VS − VCi = exp V S − v C ( t) ￿ t RC ￿ ￿ t VS − vC (t) = [VS − VCi ] exp − RC ￿ t vC (t) = VS − [VS − VCi ] exp − RC EE 442 ￿ ￿ RC and RL transients – 8 ￿ t vC (t) = VS − [VS − VCi ] exp − RC ￿ The quantitative result matches the earlier qualitative argument exactly. At...
View Full Document

Ask a homework question - tutors are online