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3 years ago

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A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where

t is in seconds and I is in amperes. The capacitor initially has no charge.
(a) Determine the voltage across the inductor as a function of time. mV
(b) Determine the voltage across the capacitor as a function of time. (V/s2) t2
(c) Determine the time when the energy stored in the capacitor first exceeds that in the inductor.

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

Explanation:

Inductance L = 1.4 x 10⁻³ H

Capacitance C = 1 x 10⁻⁶ F

a )

current I = 14 .0 t

dI / dt = 14

voltage across inductor

= L dI / dt

= 1.4 x 10⁻³ x 14

= 19.6 x 10⁻³ V

= 19.6 mV

It does not depend upon time because it is constant at 19.6 mV.

b )

Voltage across capacitor

V = ∫ dq / C

= 1 / C ∫ I dt

= 1 / C ∫ 14 t dt

1 / C x 14 t² / 2

= 7 t² / C

= 7 t² / 1 x 10⁻⁶

c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

energy stored in inductor

= 1/2 L I²

energy stored in capacitor

= 1/2 CV²

After time t

1/2 L I² = 1/2 CV²

L I² = CV²

L x ( 14 t )² = C x ( 7 t² / C )²

L x 196 t² = 49 t⁴ / C

t² = CL x 196 / 49

t = 74.8 μ s

After 74.8 μ s energy stored in capacitor exceeds that of inductor.

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