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velikii [3]
2 years ago
6

An L−R−C series circuit has 91.5 Ω , and the amplitude of the voltage across the resistor is 36.0 V . What is the average power

delivered by the source? Express your answer with the appropriate units.
Physics
1 answer:
sesenic [268]2 years ago
3 0

Answer:

14.16 W

Explanation:

We have given impedance Z=91.5 ohm

Voltage across the resistor = 36 V

We have to find the power delivered by the source

We know that according to telegens theorem , the absorbed = power delivered

So the voltage across the resistor = 36 V

Current in the RLC circuit i=\frac{V}{Z}=\frac{36}{91.5}=0.3934A this current will be same in the whole circuit as it is a series RLC circuit

Now the power absorbed by the resistor P=i^2R=0.3934^2\times 91.5=14.16W

As the power absorbed = power delivered

So power delivered by the source =14.16 W

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Read 2 more answers
You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of
alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0
3 years ago
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