Question

4) 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 1H inductor (huge!) to it.(a) how long will it take for the supply to go into current-limit

Answer 1

Answer:

0.2 seconds

Explanation:

Given data  :

Output Data = 1 ampere

supply voltage = 5 volts

inductor = 1 H

A) calculate how long before the supply goes into current limit

we apply this relationship used for determining output voltage

V = $L \frac{di}{dt}$ ------- ( 1 )

L = 1 H

di = 1 ampere

dt = ?

V = 5 volts

considering equation 1 above

dt ( time before supply goes into current limit ) =  1 / 5 = 0.2 seconds

Answer 2

it will take 0.2 seconds to go into the current limit.

Time to reach the current limit:

Given information:

Output current I = 1 ampere

Supply voltage V = 5 volts

inductor L = 1 H

The relation between current, voltage, and inductance is given below :

V = L (di/dt)

The voltage drop across the inductor is directly proportional to the rate of change of current in the inductor circuit.

Rearranging the above equation we get:

Vdt = Ldi

According to the question:

di = 1 A

so the time taken is:

5dt = 1×1

dt = 1/5

dt = 0.2 s

So it takes 0.2 second to reach the limit.

Learn more about inductance:

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