Answer:
a. 149.74 KJ/KG
b. 97.9%
c. 0.81 kJ/kg K
Explanation:
How much power do a capacitor and inductor dissipate? Assume the capacitor/inductor have no parasitic resistance (no resistor in
1 answer:
You might be interested in
Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ =
Hence,
Plug in the values for point P:
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3