Calculating the Steady State Value of Vc After Opening a Switch in a RC Circuit

[compEng]

Please Help me with this problem or with understanding it. Thank you!

Homework Statement

What is the stead-state value of V_c after the switch opens? Determine how long it takes after the switch opens before V_c is within 1 percent of its stead-state value.

Homework Equations

V_c = V_ie^-t/RC

The Attempt at a Solution

 

[L2E]

It doesn't look like you've attempted the solution. but I'll give you some tips.
when the system is in a steady state the capacitor acts as an open circuit (ie: all the current goes through the 1k resistor.)

what is the voltage across the 1k resistor at steady state: 1x10^3 x 10x10^-3
what do you know about elements connected in parallel.

I've made this part really obvious for you. finish the first part and attempt the second part and I'll give you some more help if you need it.

Good luck

 

[compEng]

okay i think I figured it out.

Because the capitor becomes an open circuit, and because it is in parrelel with the resistor it begins to imitate a Norton Equivilant circuit. and Vc=IR

V_c= 10 * 10^-3 * 10³ = 10

thank you!