Question

( www R A resistor and inductor are connected to a 9,0 V battery by a switch as shown. The moment the switch is closed, current flows through the circuit. The resistor has a resistance of R = 220, and the inductor has an inductance of L=85mH. Randomized Variables R=22002 1. = 85 mH L 9.0V 00000000 4 20% Part (a) At time t=0 the switch is closed and current flows through the circuit. Th current increases with time and eventually reaches a steady state value of imar. Calculate the maximum current imar in units of milliamps. 1 a 20% Part (b) Calculate the time constant, t, of the circuit, in seconds. A 20% Part (c) Write an equation that relates the current as a function of time i(t) to the maximum current, imax. Express the equation in terms of imax and a, where a=-t/t. m 20% Part (d) Determine the time, in seconds, at which the current has a value of i(t50) 50% of imax 20% Part (C) Determine the time, in seconds, at which the current has a value of iſt99) = 99% of imax-

Answer 1

The maximum current in the RL circuit is 40.9 mA. The time constant is 0.386 s. The equations that relate the current to time and how to calculate the time at specific current levels are provided.

Step-by-step explanation:

To calculate the maximum current imax in an RL circuit, we can use the equation imax = ε/R, where ε is the voltage of the battery and R is the resistance. In this case, ε = 9.0 V and R = 220 Ω. Substituting these values into the equation, we get imax = 9.0 V / 220 Ω = 0.0409 A. To convert this to milliamps, we multiply by 1000, giving us a maximum current of 40.9 mA.

The time constant (t) of an RL circuit is given by the equation t = L/R, where L is the inductance and R is the resistance. In this case, L = 85 mH and R = 220 Ω. Substituting these values, we get t = (85 mH)/(220 Ω) = 0.386 s.

The equation that relates the current as a function of time i(t) to the maximum current imax can be written as i(t) = imax(1 - e^(-t/t)), where t is the time in seconds. This equation shows how the current in the circuit changes over time.

To calculate the time at which the current has a value of 50% of imax, we substitute this value into the equation i(t) = imax(1 - e^(-t/t)) and solve for t. Similarly, to calculate the time at which the current has a value of 99% of imax, we substitute this value into the equation and solve for t. These values will give us the times in seconds when the current reaches the specified percentages of imax.