) a 45-mh ideal inductor is connected in series with a 60-ω resistor through an ideal 15-v dc power supply and an open switch. if the switch is closed at time t = 0 s, what is the current 7.0 ms later?a.250 ma

b.850 ma

c.550 ma

d.280 ma

e.650 ma

the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).

To find the current through the circuit 7.0 ms after the switch is closed, we can use the concept of an RL circuit. The current in an RL circuit follows an exponential growth equation, given by:

I(t) = (V/R)(1 - e^(-t/τ))

Where:

I(t) is the current at time t.

V is the voltage from the power supply (15 V in this case).

R is the resistance (60 Ω).

τ (tau) is the time constant of the circuit, given by L/R, where L is the inductance (45 mH = 0.045 H).

First, calculate the time constant τ:

τ = L/R = 0.045 H / 60 Ω = 0.00075 s.

Now, plug in the values into the equation to find I(7.0 ms):

t = 7.0 ms = 0.007 s.

I(0.007 s) = (15 V / 60 Ω) * (1 - e^(-0.007 s / 0.00075 s))

I(0.007 s) = (0.25 A) * (1 - e^(-9.333...))

Now, calculate the current:

I(0.007 s) ≈ (0.25 A) * (1 - e^(-9.333...))

I(0.007 s) ≈ (0.25 A) * (1 - 0.0000962) [Using e^(-9.333...) ≈ 0.0000962]

I(0.007 s) ≈ (0.25 A) * (0.9999038)

I(0.007 s) ≈ 0.24998 A

I(0.007 s) ≈ 250 mA

So, the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).

Learn more about RL circuit here:

brainly.com/question/17050299

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Initial current = 0

Final current = (15 V) / (60 ohms) = 0.25 Ampere

Current along the way = 0.25 · (1 - e^- time / time-constant)

"time-constant" = L/R = (0.045 / 60) =  7.5 x 10⁻⁴ second

Current = 0.25 · (1 - e^-10,000t/7.5)

When t = 7 ms,

Current = 0.25 · ( 1 - e^-70/7.5)

Current = 0.25 · (1 - e^-9.33)

Current = 0.25 · (1 - 8.84 x 10⁻⁵)

Current = 0.25 · (0.9999)

Current = so close to 250 mA that you can't tell the difference.

The reason is that 7.0 mS is 9.3 time-constants, and during EVERY time-constant, the current grows by 37% of the distance it still has left to go. So after 9.3 of these, it's practically AT the target.

I have a feeling that the time in the question is SUPPOSED TO BE 7 microseconds.  If that's true, then

Current = 0.25 · (1 - e^-[ 7 x 10⁻⁶ / 7.5 x 10⁻⁴ ]

Current = 0.25 · (1 - e^-0.00933)

Current = 0.25 · (1 - 0.9907)

Current = 0.25 · (0.0093)

Current = 2.32 mA  ?

No, that can't be it either.

Well !  Now, I'm going to determine the true and correct final answer in the only cheap and sleazy way I have left ... by looking at all the choices offered, and eliminating the absurd ones.

The effect of an inductor in the circuit is to resist any change in current.  The final current in this circuit is when it's not trying to change any more.  So the final current is just the battery with a resistor across it ... (12 V) / (60 ohms).  That's 0.25 Ampere, or 250 mA.  The current starts at zero when the switch closes, and it builds up and builds up to 250 mA.  It's never more than 250 mA.  

So look at the choices !  The only one that not more than 250 mA is choice-A .

THAT has to be it.  7.0 mS is a no-brainer.  It's 9.3 time-constants after the switch closes, the current has built up to 99.99% of its final value by then, it's not really trying to change much any more, the inductor is just about finished having any effect on the current, and the current is essentially at its final value of 250 mA.  The action is all over.

Now, I fully realize that Mister "Rishwait" is a bot and all, and nobody really needs the answer to this question.  But every cloud has a silver lining.  It's a numskull question, but it earned me 10 points, and it's been a truly fascinating trip down Memory Lane.