Question

The switch S is closed at t = 0 (assume that the battery voltage remains constant at 10V and the resistance of the inductor is negligible). Calculate the voltage across each resistor a very long time after the switch has been closed and all currents and voltages reached steady values.

A very long time after t = 0, when all currents and voltages in the circuit have settled down to steady values, the switch is opened again. What will be the current flowing through the 100-Ohm resistor 0.02 seconds after the switch is opened Make sure that you specific its direction (UP or DOWN)?

Answer 1

V across 10Ω resistor = 10V, V across 100Ω resistor = 0V, I across 100Ω resisitor = 0.632A, with the direction being UP.

Step-by-step explanation:

A long time after the switch is closed, the current through the inductance does not change . Voltage across the inductor is V(2H)=L x (dI)/(dt)=0 . It can be considered as zero resistance. It is in parallel with 100Ω resistor, so resistance of the combination will be zero. Now the circuit consists of a battery of emf 10V and a resistance 10Ω .

Current through the circuit is I = (10)/(10) = 1A

Voltage across the 10Ω resistor is V(10Ω) = 10V

Voltage across the inductor is V(2H) = 0V

Voltage across the 100Ω resistor is V(100Ω) = 0V

Current through 10Ω is I(10Ω) = 1A

Current through inductor is I(2H) = 1A

Current through 100Ω resistor is I(100Ω) = 0A

2)

After currents and voltages reached steady state again, the switch is opened again.

Now the circuit consists of 100Ω resistor and 2H inductor in series. At time t=0 , the current through the circuit is I₀ = 1A ,

At time t=0 , the current through the 2H inductor is I(2H) = 1A and is flowing down the inductor.

Current through the 100Ω resistor after time t is I(100Ω)=I₀ x (1-e^{-Rt/L})

I(100Ω) = 1(1-e^{-100*0.02/2}) = 0.632A

Direction of current in 100Ω resistor is UP.