Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
150
A
Explanation:
V
s
V
p
=
N
s
N
p
(
1
)
N
refers to the number of turns
V
is voltage
s
and
p
refer to the secondary and primary coil.
From the conservation of energy we get:
V
p
I
p
=
V
s
I
s
(
2
)
From
(
1
)
:
V
s
V
p
=
900
00
3
00
=
300
∴
V
s
=
300
V
p
Substituting for
V
s
into
(
2
)
⇒
V
p
I
p
=
300
V
p
×
0.5
∴
I
p
=
150
A
Seems a big current.
Answer:
The MATLAB Code for this PI Controller will be:
Kp = 350;
Ki = 300;
Kd = 50;
C = pid(Kp,Ki,Kd)
T = feedback(C*P,1);
t = 0:0.01:2;
step(T,t)
Explanation:
When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.
Obtain an open-loop response and determine what needs to be improved
Add a proportional control to improve the rise time
Add a derivative control to reduce the overshoot
Add an integral control to reduce the steady-state error
Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.
The further explanation is attached in the Word File.
Answer:
Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.
Explanation:
Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.
Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.
Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality = 
exit quality = 
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
