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Montano1993 [528]
2 years ago
12

An l-c cirucit with a 70 mh inductor and a ,54 F capacitor oscillates. The maximum charge on the capacitor is 11.5 C. What are t

he oscillation frequency and the maximum current in this circuit
Physics
1 answer:
Oxana [17]2 years ago
5 0

This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

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Answer:

12°F

Explanation:

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Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

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2 years ago
At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.
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3 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

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h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

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3 years ago
What is the net force on a bathroom scale when a 110 pound person stands on it? How do you know?
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8 0
2 years ago
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
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