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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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Answer:

A., B., and C.

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An Ohmic material is a material that obeys Ohm's Law, V = IR.

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Ohm's law states that the voltage across an electrical object is proportional to the current flowing through it, with the constant of proportionality being Resistance, R (in Ohm's).

The only Non-Ohmic material is the semiconductor, as semiconductors do not obey Ohm's law.

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4 0
2 years ago
Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

Where Ep = Gravitational potential, m = mass of the diver,h = Height.

But,

W = mg.................... Equation 2

Where W = weight of the diver.

Substitute equation 2 into equation 1

Ep = Wh

Make W the subject of the equation

W = Ep/h................... Equation 3

Given: Ep = 25000 J, h = 44 m

Substitute into equation 3

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W = 568.18 N.

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5 0
3 years ago
Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on
lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0
3 years ago
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