Answer:
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answer : NOR1(q_) wave is complementary to NOR2(q)
Explanation:
Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_
Initial state is unknown i.e q = 0 and q_= 1
from the diagram the waveform reset and set
= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while
from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )
From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.
From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table
also from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.
since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)
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Answer:
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R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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Answer:
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Explanation:
