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)?

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

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.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.