Question

A common demonstration of inductance employs a circuit such as the one shown in the figure (Figure 1) . Switch S is closed, and the light bulb (represented by resistance R1) just barely glows. After a period of time, switch S is opened, and the bulb lights up brightly for a short period of time. To understand this effect, think of an inductor as a device that imparts an "inertia" to the current, preventing a discontinuous change in the current through it.

A)

Derive, as explicit function of time, expression for i1 (the current through the light bulb) after switch S is closed.

Express your answer in terms of the variables E, R1, R2, L, and t.

B)

Derive, as explicit function of time, expression for i2 (the current through the resistor R2) after switch Sis closed.

Express your answer in terms of the variables E, R1, R2, L, and t.

C)

After a long period of time, the current i1 reaches their steady-state value. Obtain expression for this steady-state currents.

Express your answer in terms of the variables E, R1, R2, L, and t.

D)

After a long period of time, the current i2 reaches their steady-state value. Obtain expression for this steady-state currents.

Express your answer in terms of the variables E, R1, R2, L, and t.

E)

Switch S is now opened. Obtain an expression for the current through the light bulb inductor as an explicit function of time.

Express your answer in terms of the variables E, R1, R2, L, and t.

F)

Switch S is now opened. Obtain an expression for the current through the inductor as an explicit function of time.

Express your answer in terms of the variables E, R1, R2, L, and t.

Figure 1 ▼of! ▼ of 1 2

Answer 1

5 Take the inverse Laplace transformation. i,(S)(R, +SL) If the circuit reaches the steady state value, the voltage drop at the inductor is zero