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Norma-Jean [14]
3 years ago
5

The power supplies in 1206 Mudd feature an internal protection mechanism that limits the output current to 1 ampere (oh no, not

that again). Ideally, the protection circuit engages instantaneously when an over-current fault is detected, then disengages instantaneously once the fault condition is removed. While in the current limited state, the supply acts like an ideal current source. Now suppose you set the supply to 5 volts then connect a 1Hinductor (huge!) to it
(a) how long will it take for the supply to go into current-limit?
(b) what is the supply (and inductor) voltage once current-limit?
(c) plot the current through the inductor from the time it is connected to the supply until the time calculated in (a)
(d) Is there some value of inductance you could attach to the power supply that would not cause it to go into current-limit? lf so, what is it?
Physics
1 answer:
Art [367]3 years ago
7 0

Answer:

thanks for the points

Explanation:

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Answer:

First.

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What is Triangle ill send a picture​
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Wittaler [7]

Dry air adjacent to the cloud is entrained air is drier than the air within a cloud. The evaporation occurs in the cloud which cools the air.  The cooling of air increases its density and creates a downdraft.

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5 0
1 year ago
A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.
inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure P=1.62\times10^{6}\ N/m^2

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

Q=Av

v=\dfrac{Q}{A}

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}

v=12.44\ m/s

We need to calculate the Reynolds number

Using formula of the Reynolds number

n_{R}=\dfrac{2\rho\times v\times r}{\eta}

Where, \eta =viscosity of fluid

\rho =density of fluid

Put the value into the formula

n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}

n_{R}=7.945\times10^{5}

Hence, The Reynolds numbers for flow in the fire hose.

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