Question

1. 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. (5 points)1. 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. (5 points)

10 22 1. 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. (5 points) 10 V 100 22 2. 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)? (5 points)

Answer 1

1)

A long time after the switch is closed, the current through the inductance does not change . Voltage across the inductor is $V_{2H}=L\frac{dI}{dt}=0$ . It can be considered as zero resistance.It is in parallel with $100\,\Omega$ resistor, so resistance of the combination will be zero. Now the circuit consists of a battery of emf
OV
and a resistance $10\,\Omega$ .

Current through the circuit is $I=\frac{10}{10}=1\,A$

Voltage across the $10\,\Omega$ resistor is $V_{10\,\Omega}=10\,V$

Voltage across the inductor is $V_{2H}=0\,V$

Voltage across the $100\,\Omega$ resistor is $V_{100\,\Omega}=0\,V$

Current through $10\,\Omega$ resistor is $I_{10\,\Omega}=1\,A$

Current through inductor is $I_{2H}=1\,A$

Current through $100\,\Omega$ resistor is $I_{100\,\Omega}=0\,A$

2)

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

Now the circuit consists of $100\,\Omega$ resistor and inductor in series. At time $t=0$ , the current through the circuit is $I_0=1\,A$ ,

At time $t=0$ , the current through the   inductor is $I_{2H}=1\,A$ and is flowing down the inductor.

Current through the $100\,\Omega$ resistor after time is $I_{100\,\Omega}=I_0(1-e^{-Rt/L})$

$I_{100\,\Omega}=1(1-e^{-100*0.02/2})=0.632\,A$

Direction of current in $100\,\Omega$ resistor is UP.