Answer: E. 96. What is correct inverse Laplace transform VC(t) for the function VC(s)=(- 0.6s/s2+4)

 

96. What is correct inverse Laplace transform V_C(t) for the function V_C(s)=(- 0.6s/s²+4)

A. – 0.6 cos2t B.0.6cos2t C.- 0.6sin2t D.0.6sin2t E.0.6e^2t

 

Solution:

Answer: A.

 

97. When we transform the circuit from time domain into S-domain the admittance of inductor is

A.Y_L(s)=Ls B. Y_L(s)=Ls with i_L(0)=0 C. Y_L(s)=1/Ls D. Y_L(s)=1/Ls with i_L(0)=0

E. Y_L(s)=1/L

 

Solution:

Z_L=Ls with i_L(0)=0; Y_L=1/Z_L=1/Ls with i_L(0)=0;

Answer: D.

97. Define impedance of the circuit in the s-domain respectively points A and B (fig.11).

A.Z(s)=RCs+1 B.Z(s)=Cs+2R C.Z(s)=RCs+R D.Z(s)= R(RCs+1)/(RCs+2)

E. Z(s)=R(RCs+2)/(RCs+1) Fig.11

Solution:

Answer: E.

 

98. Find the mistake in process of transformation of time-domain circuit in s-domain circuit.

A. wrong representation of the capacitor impedance

B. wrong representation of the inductor impedance

C. wrong representation of the capacitor initial condition current source

D. wrong representation of the inductor initial condition current source

E. there is no mistake in process of circuit’s conversion

 

 

 

 

Solution:

I(t)=I(s);

Z_R(s)=R;

Z_C(s)=1/Cs;

I_C(s)=CsV_C(s) – CV_C(0);

Z_L(s)=Ls;

I_L(s)=(1/Ls)*V_L(s)+ i_L(0)/s;

Wrong representation of the inductor initial condition current source.

Answer: D.

 

99. The time domain equation is 10i₁+0.02(di₁/dt)-0.02(di₂/dt)=100. What is the view of the equation in the s-domain?

A. (10+0.02s)I₂(s)-0.02sI₁(s)=100/s B. (10+0.02s)I₁(s)-0.02sI₂(s)=100

C. (10+0.02s)I₁(s)-0.02sI₂(s)=100/s D. (10+0.04s)I₁(s)-0.04sI₂(s)=100/s

E. 0.02sI₁(s)-0.02sI₂(s)=100/s

 

Solution:

time domain	s-domain
f(t)	F(s)
df(t)/dt	sF(s)-f(0-)
	100/s

t-domain:

 

10i₁+0.02(di₁/dt)-0.02(di₂/dt)=100

s-domain:

10I₁+0.02sI₁ – 0.02sI₂=100/s

(10+0.02s)I₁ – 0.02sI₂=100/s