Given the CR circuit below, answer the questions given a period of time. Briefly

Question

Given the CR circuit below, answer the questions given a period of time. Briefly explain your answers. Note: Before time zero, the switch is off and the capacitor is fully discharged. R2 RI (V2) Figure 1: CR Circuit 1. The switch closes at t 0. (a). What is the current (lc) through Catt 0? (b). What is the current (lR) through R1 at t 0? (c). What is the voltage drop across C at t 0 2. The switch is closed for a very long time (more than 5 time constants). (a). What is the current (Ie) through C? (b). What is the current (ln) through R1? (c). What is the voltage drop across C? 3. The switch is immediately re-opened (after the point in time of Prob.2). (a). What is the current (lc) through C? (b). What is the current (l&) through R1? (c). What is the voltage drop across C? 4. The switch is open after a very, very long time (after the point in time of Prob. 3). (a). What is the current (le) through C? (b). What is the current (k) through R1? (c). What is the voltage drop across C?

Explanation / Answer

This is an RC circuit with a step input.

Terminology used: Vc = Voltage across capacitor, VR = Voltage across resistor, Ic = Current across capacitor, IR = Current across resistor, Vs = Source voltage = 5 V, T = time constant = Req C

Given : Vc (0-) = Voltage across capacitor at time t<0 = 0V.

The voltage across the capacitor at time t>0 is given by = Vc(t) = Vs (1 - e-t/T) .

We need to find the time constant now, Req is the thevnin equivalent of this circuit, which is 1k ohm in parallel with 1k ohm, which is 0.5k ohm. and C = 1nF, that gives Req C = 0.5k ohm x 1nF = 1 micro ohm.F

Hence Vc(t) can be written as Vc(t) = 5 (1 - e-t/1 micro ohm.F ).

and current through the capacitor can be given by: Ic = C. dVc(t)/dt = Vs/R(e-t/1 micro ohm.F ).

Now we know the voltage and current across the capacitor, we need to find the voltage and current across the resistance, As the capacitor and the resistor, are in parallel, the voltage across them is equal.

Hence VR = Vc(t) and IR = VR/R = VR/1k.ohm.

Now we can start answering the questions:

Part 1: At time t = 0

Vc(0) = 5 (1 - e-0/1 micro ohm.F ) = 0V, VR = Vc = 0V, IR = VR/R = 0mA

Part 2: At time t = 5 time constants = t/T = 5

Vc(0) = 5 (1 - e-5 ) = 5x0.993326 = 4.9663 V, VR = Vc = 4.9663 V, IR = VR/R = 4.9663 V/1k.ohm = 4.9663 mA

Part 3: Switch is reopened after 5 time constants:

After the switch is opened, this becomes an RC circuit: the voltage across capacitor and the resistor is the same and is given by: Vc = VR = V0 e-t/RC, here V0 is the inital voltage across the capacitor, which is the final voltage across the capacitor after 5 time constants = V0 = 4.9663 V

Then Vc = VR = V0 e-t/RC = 4.9663 ( e-t/RC ) immediately after the switch is open, Vc = VR   = 4.9663 ( e-0/RC ) = 4.9663 V and hence IR = VR/R = 4.9663 V/1k.ohm = 4.9663 mA

Part 4: Switch is opened and left fot a very long time = t = infinity

Then Vc = VR = V0 e-t/RC = 4.9663 ( e-t/RC ) after a very long time, Vc = VR   = 4.9663 ( e-infinity ) = 0 V and hence IR = VR/R = 4.9663 V/1k.ohm = 0 mA

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.