] In the circuit diagram below, v(t) is a 4Vpp triangular wave at a frequency of

Question

]   In the circuit diagram below, v(t) is a 4Vpp triangular wave at a frequency of 5000Hz. Resistor R is 1K? and inductor L is 5mh. The current in the circuit, the current flowing through the inductor, can be found by applying Ohm

In the circuit diagram below, v(t) is a 4Vpp triangular wave at a frequency of 5000Hz. Resistor R is 1K? and inductor L is 5mh. The current in the circuit, the current flowing through the inductor, can be found by applying Ohm's Law of v(t)/R. With this periodic waveform, calculate and show the voltage, vL, the voltage across the inductor. I am looking for solutions to this problem. Very confused. Need to use differential equations to solve.

Explanation / Answer

let i curent flow through circuit :::

Apply kvl :::

iR + Ldi/dt = v(t)

Ldi/dt +iR = v(t) ................(1)

equation (1) have two type of solution   ,,, one solution is called homogeneous solution ,,, ad another is called particular solution ....

let first find the homogeneous solution of   Ldi/dt +iR =0

                                                                    di/dt    +i*(R/L) =0

                                                                    

                              ih(t) = homogeneous solution = Ae^(-Rt/L)

for particular solution ::: let    ip(t) = particular solution = k*v(t) ............(2)

this ip(t) will satisfy the (1) ... hence put ip(t) in (1) ans solve for k ....

we get    k = 1/R

hence ip(t) = V(t)/R

i(t) = total solution = ip(t) + ih(t)

   = Ae^(-Rt/L) + V(t)/R

hence   i(t) = Ae^(-Rt/L) + V(t)/R

constant A can be found from initila condition :::

at t =0 , i(t) =0 , hence   A = V(0)/R   = 4Vpp/(2R) = 2Vpp/R

i(t) = [V(0)/R]e^(-Rt/L) + V(t)/R

i(t)   = V(t)/R     , at t = infinity

         = 4Vpp/1

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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