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krok68 [10]
3 years ago
6

A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 1 H and a resistance of 1.4

kΩ. Switch S1 is suddenly closed at t = 0. Find the time needed for the current to reach a fraction f = 0.6 of its maximum value.
Physics
1 answer:
shtirl [24]3 years ago
8 0

Answer:

\tau \approx 7.14 \times 10^{-4}s \approx0.714ms

Explanation:

In a LC circuit The time constant τ is the time necessary for 60% of the total current (maximum current), pass through the inductor after a direct voltage source has been connected to it. The time constant can be calculated as follows:

\tau =\frac{L}{R}

Therefore, the time needed for the current to reach a fraction f = 0.6(60%) of its maximum value is:

\tau =\frac{1}{1.4\times 10^{3}} =7.142857143 \times 10^{-4} \approx7.14 \times 10^{-4}s

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Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
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<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

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Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

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Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

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