A 7.35 H inductor with negligible resistance is placed in series with a 14.1 V battery, a 3.00 ? resistor, and a switch. The switch is close

Question

A 7.35 H inductor with negligible resistance is placed in series with a 14.1 V battery, a 3.00 ? resistor, and a switch. The switch is closed at time t = 0 seconds.

(a) Calculate the initial current at t = 0 seconds.

(b) Calculate the current as time approaches infinity.

(c) Calculate the current at a time of 6.25 s.

(d) Determine how long it takes for the current to reach half of its maximum.

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Neala 3 weeks 2021-09-03T17:09:44+00:00 1 Answers 0 views 0

Answers ( )

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    2021-09-03T17:10:44+00:00

    Answer:

    (a). The initial current at t=0 is zero.

    (b). The current as time approaches infinity is 4.7 A

    (c). The current at a time of 6.25 s is 4.33 A.

    (d). The time is 1.69 sec.

    Explanation:

    Given that,

    Inductance = 7.35 H

    Voltage = 14.1 V

    Resistance= 3.00 ohm

    (a). We need to calculate the initial current at t = 0

    Using formula of current

    I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})

    Put the value into the formula

    I=(\dfrac{E}{R})(1-e^{\dfrac{-0\timesR}{L}})

    I=(\dfrac{E}{R})(1-1)

    I=0

    (b). We need to calculate the current as time approaches infinity.

    Using formula of current

    I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})

    Put the value into the formula

    I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-\infty\times3.00}{7.35}})

    I=\dfrac{14.1}{3.00}(1-0)

    I=4.7\ A

    (c).  We need to calculate the current at a time of 6.25 s

    Using formula of current

    I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})

    Put the value into the formula

    I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-6.25\times3.00}{7.35}})

    I=4.33\ A

    (d). We need to calculate the time

    Using  formula of current

    I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})

    Put the value into the formula

    2.35=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-t\times3.00}{7.35}})

    ln(0.5)=(\dfrac{-t\times3.00}{7.35}})

    t=\dfrac{(\ln(2)\times7.35)}{3.00}

    t=1.69\ s

    Hence, (a). The initial current at t=0 is zero.

    (b). The current as time approaches infinity is 4.7 A

    (c). The current at a time of 6.25 s is 4.33 A.

    (d). The time is 1.69 sec.

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