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Lady bird [3.3K]
3 years ago
7

In an oscillating L C circuit, the maximum charge on the capacitor is 1.5 × 10 − 6 C and the maximum current through the inducto

r is 5.5 mA.
(a) What is the period of the oscillations?
(b) How much time elapses between an instant when the capacitor is uncharged and the next instant when it is fully charged?
Physics
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

T= 1.71×10^{-3} sec= 1.71 mili sec

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

Explanation:

First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)=  5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C

a) I(t) = -ωqosin(ωt+φ)

⇒Io= ωqo

⇒ω= Io/qo

also we know that T= 2π/ω

⇒T= 2\pi\frac{q_0}{I_0}

now putting the values we get

= 2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}

= 1.71×10^{-3} sec

b) note that the time t_{fc} it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is

t_{fc}= \frac{T}{4}

t_{fc}= \frac{ 1.71\times10^{-3} }{4}

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

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The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.

differentiating the equation h.

dh/dx = h' = -2x + 3 = 0

Solving for x;

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Substituting into the function h;

h max = -x^2+3x+3

h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =

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Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

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y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

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