Answer:
Option B is right answer.
Yes, because both figures are rectangles and all rectangles are similar.
<u>Answer</u>-
<em>The total capacitive reactance of the circuit is </em>442.10<em> Ω.</em>
<u>Solution-</u>
Three capacitors, a 12 mF, a 20 mF, and a 30 mF, are connected in series to a 60 Hz source.
The effective/equivalent capacitance of the circuit is,
When capacitors are joined in series, the total value of capacitance in the circuit is equal to the reciprocal of the sum of the reciprocals of capacitance of all the individual capacitors.
We know that,
Where,
= Capacitive Reactance (in Ω)
f = frequency (in Hz)
C = Capacitance (in F)
Here given,
= ??
f = 60 Hz
C = 6 mF = F
Putting the values in the formula,
Therefore, the total capacitive reactance of the circuit is 442.10 Ω.
Answer: 113.04 (option 3)
Step-by-step explanation: Area = pi · radius²
If we take the Pythagorean identity identity sin^2 x + cos^2 x = 1 then
<span>(cos^2 x + sin^2 x) / (cot^2 x - csc^2 x)
The numerator becomes 1 since addition order matters not.
1 / </span>(cot^2 x - csc^2 x)
If we factor the denominator out a negative
1 / -(<span>csc^2 x - cot^2 x)
Consider </span><span>sin^2 x + cos^2 x = 1. Divide both sides by sin^2 x to get
1 + cot^2 x = csc^2 x
Subtract both sides by cot^2 x to get 1 = csc^2 x - cot^2 x.
Replace the denominator
1 / -(1) = -1
For cos</span>^2 θ / sin^2 θ + csc θ sin θ, we use cscθ = 1/sinθ and cosθ/sinθ = cotθ so
= cos^2 θ / sin^2 θ + 1
= cot^2 θ + 1
We use 1 + cot^2 <span>θ = csc^2 </span>θ to simplify this to
= csc^2 θ
Answers: -1
csc^2 θ