Physics, 08.04.2020 18:27 simbamillions6899

What is umax, c, the value of the maximum energy stored in the capacitor during one cycle?

Answers: 3

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Answer from: larenhemmings

1) 0.266 H

2) 0.040 J

3) $0.308\Omega$

Explanation:

The diagram of the circuit is missing: find it in attachment.

1) What is L, the value of the inductance of the circuit?

For an RLC circuit, the resonant angular frequency is given by:

$\omega=\frac{1}{\sqrt{LC}}$

where

L is the inductance of the circuit

C is the capacitance of the circuit

In this problem, we have:

$\omega=164 rad/s$ is the angular frequency of the generator, at which the circuit is in resonance

$C=140 \mu F = 140\cdot 10^{-6}F$ is the capacitance

Therefore, solving for L, we find the inductance of the circuit:

$L=\frac{1}{\omega^2 C}=\frac{1}{(164)^2(140\cdot 10^{-6})}=0.266 H$

2) What is umax,c, the value of the maximum energy stored in the capacitor during one cycle

The maximum energy stored in a capacitor is given by the equation

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the maximum potential difference across the capacitor

Here we have:

$C=140 \mu F = 140\cdot 10^{-6}F$ is the capacitance

V=24 V is the maximum voltage across the capacitor, which is the emf of the generator

Substituting, we find:

$U=\frac{1}{2}(140\cdot 10^{-6})(24)^2=0.040 J$

3) What is R, the value of the resistance of the circuit?

Here we want to find the resistance of the circuit.

The resistance of the circuit can be found by using Ohm's Law:

V=RI

where

V is the maximum voltage

R is the resistance

I is the maximum current

Here we have:

V = 24 V is the maximum voltage provided by the generator

I = 0.78 A is the maximum current in the circuit

Solving for R, we find the resistance:

$R=\frac{V}{I}=\frac{0.24}{0.78}=0.308\Omega$

What is umax,c, the value of the maximum energy stored in the capacitor during one cycle?

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

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