Answer:
For any string, we use 
Explanation:
The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.
Here are the cases:
- Consider any string a i b j c k in the language. If i = 1 or i > 2, we take
and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
- For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
- Finally, for the case i = 0, we take , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
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Answer:
The value of v2 in each case is:
A) V2=3v for only Vs1
B) V2=2v for only Vs2
C) V2=5v for both Vs1 and Vs2
Explanation:
In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.
Also, what the problem asks is the value V2 in each case, where:
If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.
In the first case we can use an equivalent resistance between R2 and R3:
And
In the second case we can use an equivalent resistance between R2 and (R1+R4):
And
If we consider both batteries:
Answer:
Letter D
Explanation:
AutoCAD provides eleven different ways to create arcs. The different options are used based on the geometry conditions of the design. To create an arc, you can specify various combinations of center, endpoint, start point, radius, angle, chord length, and direction values.
The equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.
<h3>
What is resistance?</h3>
Resistance is the obstruction offered whenever the current is flowing through the circuit.
So the equivalent resistance is when three resistances are connected in series. When the resistances are connected in series then the voltage is different and the current remain same in each resistance.
V eq = V₁ + V₂ + V₃
IReq = IR₁ + IR₂ + IR₃
Req = R₁ + R₂ + R₃
Therefore the equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.
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