Loading....
Coupon Accepted Successfully!

 

Solved Problems-17

Problems-17
The figure shows a parallel RLC circuit fed from a dc current source through a switch. The circuit elements are R = 400Ω, L = 25mH, C = 25nF. The source current is 24mA. The switch which has been in closed position for a long time is opened at t = 0.
 
Description: Description: 4070.png
  1. What is the initial value of current iL (i.e., at t = 0)?
  2. What is the initial value of voltage across L at t = 0?
  3. What is the expression for current through inductance, capacitance and resistance?
  4. What is the final value of iL?
  5. What happens to iL(t) if R is increased from 400 to 625Ω? Assume that initial energy is zero.
Solution
Applying KCL for the node, we get,
 
Description: Description: 8945.png
 
Taking Laplace transform,
 
Description: Description: 8955.png
⇒ Description: Description: 8961.png
 
Substituting the values,
Description: Description: 8967.png
 
Taking inverse Laplace transform,
 
Description: Description: 8973.png
 
Also, the current through the inductor,
Description: Description: 8983.png
 
Taking inverse Laplace transform,
Description: Description: 8993.png
  1. At t = 0, we get, Description: Description: 8999.png
  2. At t = 0, we get, Description: Description: 9005.png
  3. Current through inductance
     
    Description: Description: 9011.png 
     
    Current through capacitance,
     
    Description: Description: 9019.png
     
    Current through resistance,
     
    Description: Description: 9026.png
  4. At t = ∞, the final value of iL is, Description: Description: 9032.png
  5. Putting the value of resistance R = 625Ω in the expression of iL, we get,
     
    Description: Description: 9038.png
Taking inverse Laplace transform and simplifying, we get,
 
Description: Description: 9044.png 
 
Here, with R = 400Ω, the circuit was in overdamped condition. As the value of resistance is increased to 625 Ω, the circuit becomes underdamped.
 




Test Your Skills Now!
Take a Quiz now
Reviewer Name