Consider a RL Circuit with L = 2.0 H and time constant of 0.2 s. The battery vol

Question

Consider a RL Circuit with L = 2.0 H and time constant of 0.2 s. The battery voltage is 12 V. Calculate the resistance R. After the switch is closed, what is the maximum current that passes through the resistor? At one time constant after the switch is closed, calculate the currant through the resistor, the voltage across the resistor and tire energy stored in the inductor. the switch replaced opened and the battery replaced by an AC generator with output voltage Delta v = (100 V) sin (3pi t) The switch is then closed. Find the inductive reactance and total impedance of this circuit The current through the circuit will the voltage be in phase with lead the voltage.

Explanation / Answer

a)

T = time constant = 0.2

L/R = 0.2

2/R = 0.2

R = 10 ohm

I = Io ( 1 - e-t/T)

at t = 0

I = 0

b)

maximum current , Io = E/R = 12 / 10 = 1.2 A

I = Io ( 1 - e-t/T)

at t = T

I = (1.2) ( 1 - e-1)

I = 0.76 A

Voltage across resistor = iR = 0.76 x 10 = 7.6 volts

energy stored in indutor = E = (0.5) L i2 = (0.5) (2) (0.76)2 = 0.58 J

c)

XL = wL = 30 (3.14) (2) = 188.4 ohm

z= = sqrt(R2 + XL2) = sqrt(102 + 188.42 ) = 188.7 ohm

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