It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.
Answer:
Explanation:
D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m
ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s
We can apply the equation
Ff = W ⇒ μ*N = m*g <em>(I)</em>
then we have
N = Fc = m*ac = m*(ω²*R)
Returning to the equation <em>I</em>
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μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)
Finally
μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379
Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V
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