Question

A circuit 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

Answer 1

Answer:

Explanation:

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

i = i₀ ( 1 - $e^{\frac{-t}{\tau}$ )

Differentiating the equation on both sides

di / dt = i₀ / τ  x $e^\frac{-t}{\tau}$

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 = $1 - e^\frac{t}{\tau}$

$e^\frac{-t}{\tau}$ = .15

t / τ = ln .15

t = 74 μs