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pochemuha
3 years ago
14

Recall that the voltages VL(t) and VC(t) across the inductor and capacitor are not in phase with the respective currents IL(t) a

nd IC(t). In particular, which of the following statements is true for a sinusoidal current driver?
VL(t) and VC(t):
a) both lag their respective current
b) both lead their respective currents
c) VL(t) lags IL(t) and VC(t) leads IC(t)
d) VL(t) leads IL(t) and VC(t) lags IC(t)
Physics
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

D) VL(t) leads IL(t) and VC(t) lags IC(t)

Explanation:

This is because The phase angle between voltage and current for inductors and capacitors is 90 degrees, or radians, also, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage,

( I(t), V(t)), is zero.

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1. Straight-in spaces can leave you a safer out if
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3 years ago
Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an
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E+mc+UD-3

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7 0
3 years ago
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect, 

Point 1: 
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m 
The distance between the two points then is equal to 7.07 m.


</span>
8 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
2 years ago
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