Question

The switch in the circuit of the following figure has been opened at t= 0 Determine Vo(t) for t> 0 2 ka w 6V 3 mi

Answer 1

SOLUTION:

The given is source free at t > 0 .

At t= 0- switch is closed hence the initial Voltage in the capacitor is calculated .

And at t = INFINITE the switch is open then the circuit is in steady state hence the circuit will be source free and the steady state voltage on the capacitor will be 0 Volt.

By the 1st order transient equation of a capacitor Voltage .

Since Voltage and resistance both are in parallel hence.

Vo(t) = Vc(t).

For any quaries let me know in the comment section.

Thank you...

1. 
   At t< "switch is closedu D switch is closed 2ku M M T + UKN Velo 6 y At t = the circuit switch is closed and is in steady state open circuted capacitog is 2 kw + 1+ UKM Vilo). 6v By the voltage divides rule Vicol= 6X.uk ukt ak V, 10) = 68 Uk 6K Vc (5) = u volt Velo) = 4V
   
2. 
   the At t=0 circuit is in steady state, capacitor is open circuited switch is open 2 kr ۸۸۸ + Velo 6V Ukr + m Velo) ИК The circuit is source free, hence Velal = 0 volt Time constant (T). T=Rthc switch is open At - to Rth Ukr
   
3. 
   Rth = Time constant IT): voltage equation in ist order tika T= Rth ac Ruth = uka (= 3MF T = UK.X 3m T 12 sec T = 12 sec capacitou The transient equation for -t e T Velt) = Viro) + V. (x) + | Velola Vella +[vc (5) Veron] Velo) = 4V Velo) = ovolt T= 12 sec -t 12 Volt [4-o] e 0 + [4 - Velt) - -t/12 volt че Veltla
   
4. 
   From the circuit attro + vo UKN + Volt 3m fi from the circuit at tro Vellt) = Volt) are equal. In Parallel voltages Voit) = = Vect) Finally Volt) for too Vold) = -tli2 4 e volt 0.0833t Volt) - 4 e -tli 4 e Volt Volt) = 71 -0.0833t 4 e Volt) Volt 11