DP 8-6 Fuses are used to open a circuit when excessive current flows (Wright, 1990). One fuse is designed to open when the power absorbed by R exceeds 10 W for 0.5 s. Consider the circuit shown in Figure DP 8.6. The input is given by v_s = A[u(t) – u(t – 0.75)] V. Assume that i_L(0^–) = 0. Determine the largest value of A that will not cause the fuse to open.

Figure DP 8.6

The voltage source provides zero voltage for t < 0. SO the current through the inductor is also zero for t < 0. After t = 0, the current will build up in the inductor. Do a voltage walk around the circuit to develop the differential equation.

simplifying

For the period 0 ≤ t ≤ 0.75 s

Assume the solution of a constant force and the natural response so

The Thevenin resistance is the two resistors in series, so

So

And

Plug into the diff-eq

So

To Find B use the value at t = 0

So

Our solution is

So for the fuse to open power delivered to resistor must exceed 10 W for at least 0.50 s. We also know voltage returns to 0 at t = 0.75 s. So we must have P approach 10 W at

t = 0.25 s.

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