Question

Suppose you have the following circuit. The switch has been open for a long time, and is closed at time t = 0.

Assume V2=5V,R2=10kΩ,L2=2mH,C2=0.1μF
The second order differential equation for the voltage across the capacitor is

d^2V_c(t)/dt + a_1dV_c(t)/dt + a_2V_c = K
What is the coefficient K of the 2nd order differential equation for this circuit? Do not use scientific

Answer 1

Explanation:

derivation distribution of easily you can open the book and read the chapter then you can understand what you're asking

Answer 2

Given:

V2=5V,R2=10kΩ,L2=2mH,C2=0.1μF

The second order differential equation for the voltage across the capacitor is

d²V_c(t)/dt² + a_1dV_c(t)/dt + a_2V_c = K

To Find:

The coefficient K of the 2nd order differential equation for this circuit.

Solution:

From the circuit in the attachment , we can easily write the Loop equation using Kirchoff's Voltage Law.

• 5 - iR - Ldi/dt - Vc  - 0
• 5 - iR - Ldi/dt - Vc = 0

We know, for capacitor ,

• Vc = Q/C
• Current in the circuit = i = dQ/dt
• dVc/dt = dQ/Cdt = i/C
• i = CdVc/dt
• di/dt = Cd²Vc/dt²

Substituting i and di/dt in the voltage equation,

• 5 - RC dVc/dt - LCd²Vc/dt² - Vc = 0

Dividing by -LC,

• d²Vc/dt² +RdVc/Ldt + Vc/LC = 5/LC

Therefore K in the second order differential equation is ,

• K = 5/LC
• K = 5/2m x 0.1u = 50/2 x $10^{-9}$  = 25 x $10^{-9}$

The numerical value of  coefficient K of the 2nd order differential equation for this circuit is

25 x $10^{-9}$ .