Question

An inductor of ????=6.95 H with negligible resistance is placed in series with a ℰ=12.5 V battery, a ????=3.00 Ω resistor, and a switch. The switch is closed at time ????=0 seconds. A square loop circuit containing a battery with E M F script E, an open switch S, and inductor of inductance L, and a resistor of resistance R, all connected in series. Calculate the initial current at ????=0 seconds. ????(????=0 s)= 0 A Calculate the current as time approaches infinity. ????max= ___ A Calculate the current at a time of 1.85 s. ????(????=1.85 s)= ___A Determine how long it takes for the current to reach half of its maximum. ????= ___s

Answer 1

Answer:

a) I=0 b) 4.17V c) 0.354 A d) 14.5s

Explanation:

a) consider circuit in the attachment

i(t)= E/R (1- e^(-t/RL))

i(0)= 12.5/3×(1-e^(0/RL))

i(0)=0

b) at t⇒∞

i(∞)= 12.5/3× (1- e^(-∞/RL))

    = 4.17V

c) 1/RL= 1/(6.95×3)= 0.0479616

i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)

         = 0.354A

d) I/2= I (1- e^(-t/RL))

t= - RL ln0.5

t= - 3×6.95 × (-0.693)

t= 14.5 s