The switch in Figure P23.39 is open for t < 0 and then closed at time t = 0. Fin

Question

The switch in Figure P23.39 is open for t < 0 and then closed at time t = 0. Find the current in the inductor and the current in the switch as functions of time thereafter.

Guide for question from prof:

With regard to proceeding, I suggest the following:

0. Notation: in what follows I(t) is the total current, I1 goes through 4ohm resistor, I2 through the 8ohm resistor and inductor. Also,

1. First identify whats happening at time t=0. You know that at the moment the switch is closed dI/dt is very large, so hardly any current flows through the inductor at t=0.

2. When t goes to infinity (inf), dI//dt goes to zero (i.e. steady current).

3. Work out these special cases first and obtain values for I1(0), I2(0), I(0) as well as I1(inf), I2(inf), I(inf).

4. Now figure out the functional form that takes you from I1(0) to I1(inf), I2(0) to I2(inf), etc.

Explanation / Answer

Inductor acts as open circuit initially and short circuit in steady state,Hence.

Current at t(0);

Inductor open circuited,

Req = 4+4 = 8;

I1(0) = I(0) = V/Req = 10/8 = 1.25;

I2(0) = 0;

Current at t(inf);

Inductor short circuited.

Req = 4+ (4*8/4+8) = 6.66;

I1(inf) = I(inf) = 10/6.66 = 1.50;

I2(inf) = 1.00;

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