Acircuit consists of a 9.0 mh inductor coil, a 230 ω resistor, a 12.0 v ideal battery, and an open switch-all connected in series. the switch closes at time t = 0. what is the initial rate of change of the current - i just after the switch closes? dt l=0 a/s what is the steady-state value of the current if a long time after the switch is closed? if= at what time 185% does the current in the circuit reach 85% of its steady-state value? 185% = what is the rate of change of the current when the current in the circuit is equa dt li=1; /2 its steady-state value? di - dt li=1; /2

Explanation:

This is the case of L-R charging circuit , for which the formula is as follows

i = i₀ ( 1 - )

Differentiating the equation on both sides

di / dt = i₀ / τ  x

i is current at time t , i₀ is maximum current , τ is time constant which is equal to L / R where L is inductance and R is resistance of the circuit .

τ = L / R = 9 x 10⁻³ / 230

= 39 x 10⁻⁶ s

i₀ = 12 / 230 = 52.17 x 10⁻³

di / dt (at t is zero) = 52.17 x 10⁻³ / 39 x 10⁻⁶

= 1.33 x 10³ A / s

Time to reach 85 % of steady state or i₀

.85 =

= .15

t / τ = ln .15

t = 74 μs

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explanation: