Solutions:

Problem #1:

R = 4/11k = 365 Ohm.

Problem #2:

Current through R5: 2A

Current supplied by V1: 9A

Problem #3:

V_Th = 0.2V

I_N = 0.333mA

R_Th = R_N = 600 Ohm

With a 1k Load, VLoad = 0.125V

Problem #4:

Equivalent Resistance = 1k

(Hint: try redrawing the circuit in the shape of a Wheatstone Bridge.)

Problem #5:

for w = 2*Pi*f:

 

A: V(t) = Vs(t) R1 /( R1 + j w L)

The absolute value of V(t)/Vs(t) is: 1/Sqrt( 1 + (w L / R)^2 ). From this it can be shown

that w3dB = R / L = 1000 radians/sec = 159 Hz

The phase is:

tan( phi ) = -w L / R = -w / 1000

 

B: for w = 0, phi = 0,

for w infinity, phi = -90 degrees.

for w3dB, phi = -45 degrees.

Problem #6:

w3dB = (R1 + R2)/(R1 R2 C1) = 1.5 x 10^5 radians/sec = 23.9kHz.

(Note: the original WEB picture had incorrectly C1 = 0.01mF, it should have read: C1= 0.01uF.)

Hint: make sure your "definition" of the attenuation takes into consideration that at w = 0, Vout/Vin is not 1!