An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.
1 answer:
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
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Answer: The correct option is Option b.
Explanation:
Power is defined as the rate of work done by an object.
Mathematically,
.....(1)
And work done is the product of force exerted on the object times the displacement covered by that object.
Mathematically,
Putting this value in above equation, we get:
where,
P = power = ?W
F = Force exerted = 10N
s = Displacement = 400cm = 4m (Conversion factor: 1m = 100 cm)
t = Time taken = 8s
Putting values in above equation, we get
Hence, the correct option is Option b.