EXERCISE 1: A NON-SERIES RL CIRCUIT n the circuit shown adjacent, V-60.0 V, R -1

Question

EXERCISE 1: A NON-SERIES RL CIRCUIT n the circuit shown adjacent, V-60.0 V, R -10.0 2, R:-20.02, and L-3H Ri Immediately after the switch is closed, what is the value of the current through R? Make sure you justify your answer 1. R: 2. Immediately after the switch is closed, what is the rate of change of current through L? Also, specify the direction (up/down) that the current flows through L. Justify your answer 3. A long time after the switch is closed, what is the value of the current through R? Justify your answer 4. Once the circuit has reached a steady state (a long time after the switch was initially closed), the switch is reopened. Immediately after the switch is reopened, what is the value of the current through R2? After the switch is reopened, how much time does it take for the current in R2 to reach approximately 37% of its initial value? Justify your answers

Explanation / Answer

Only one question can be posted as per chegg guidelines

1.

a)

Immediately after switch is closed ,the inductor will act as open circuit.The Value of current through R2 is

I=60/(10+20) =2 A

b)

Voltage across Inductor is

VL =IR2=2*20 =40 V

Therefore

dI/dt =VL/L =40/3

dI/dt =13.33 A/s

c)

After long time switch is closed ,inductor will act as short circuit .Therefore no current flows through R2 is zero.

I2=0 A

d)

When switch is reopened ,inductor does not allow sudden change in current

I=60/10 =6 A

From

Time Constant

T=L/R2 =3/20 =0.15 s

I=Ioe-t/T

=>0.37 =e-t/0.15

Ln(0.37) =-t/0.15

t =0.149 s

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