Question

(i need the answer to number 4)

For the circuits shown: XSCi XSC1 XEG2 1R1 - 100mH 10kΩ J1 key-space 10k. 5090 Key-A 수10nF 50% 0

1. For circuit (a), with J1 opened:

1. Calculate the time constant of the circuit, τ₁,

τ₁ =   100 μs

2. Using a square wave with a period 20τ₁ and a 2-V_PP, plot the wave forms for channels (1), (2), and the difference between them (v_R1), and

f₁ = 1/ (20τ₁)= 500 Hz

2. For circuit (a), with J1 closed:

1. Calculate the time constant of the circuit, τ₂, when R₂ = 100%, and

τ₂ =   50 μs

f₂ = 1/ (20τ₂)= 1 kHz

  

3. For circuit (b):

1. Calculate the time constant of the circuit, τ₃, when R = 50%,

τ₃ =   20 μs

f₃ = 1/ (20τ₃)= 25 kHz

4. Calculate i_C in circuit (a) and v_L in circuit (b).

(i need the answer to number 4)

Answer 1

With J1 open rightwardsdoublearrow The circuit reduces to- To calculate Time time - constant: - Find Reg: - keep voltage source replace by their internal resistances- voltage source = short circuited. Reg rightwardsdoublearrow Across open circuited capacitor Terminal- Reg = 10 k Ohm tau_1 = R_eq C = 10^4 times 10 times 10^-9 = 10^-4 sec tau = 100 times 10^-6 tau_1 = 100 mu sec For the circuit with J1 closed rightwardsdoublearrow R_2 = 100% ie: R_2 = 10 k ohm The network can be redrawn as- Reg: - keep V_s by short circuit Therefore Reg = R1 Parallel R_2 = 10 k Ohm Parallel 10 k Ohm R_eq = 5 k Ohm Now, Time constant tau_2 = R_eg C_eq = 5 times 10^3 times 10 times 10^-9 = 5 times 10^-5 tau_2 = 50 mu sec For the circuit (b) When key is Adjusted to 50% ie: R_2 = 10 k Ohm times 50% = 5 k Ohm Now, the Time constant- tau_3 = L_eq/R_eq = 100 times 10^-3/5 times 10^+ 3 = 2 times 10^-5 tau_3 = 20 mu sec Now, V_in is set to f_3 = 1/20 tau_3 = 10^6/20 times 20 \r\n < not >