Answer:
a) I=0 b) 4.17V c) 0.354 A d) 14.5s
Explanation:
a) consider circuit in the attachment
i(t)= E/R (1- e^(-t/RL))
i(0)= 12.5/3×(1-e^(0/RL))
i(0)=0
b) at t⇒∞
i(∞)= 12.5/3× (1- e^(-∞/RL))
= 4.17V
c) 1/RL= 1/(6.95×3)= 0.0479616
i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)
= 0.354A
d) I/2= I (1- e^(-t/RL))
t= - RL ln0.5
t= - 3×6.95 × (-0.693)
t= 14.5 s
Answer:
Explanation:
relating to, measuring, or measured by the quantity of something rather than its quality.Often contrasted with qualitative.
Answer:
(a) Calculate the field, armature, and load currents versus load = 306A
(b) Determine the terminal voltage at no-load and at rated load conditions = 300V
(c) Calculate the voltage regulation of the generator. Use the no-load voltage as the base value = 6.67%
(d) Plot the terminal voltage as a function of the load. Determine the load that corresponds to a 5% voltage drop using the no-load voltage as the base = 294.74V
Explanation:
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