Answer
Put a 98 mH inductance in series with the device.
Work Step by Step
Placing an inductor in series with the device will protect it from sudden current surges. See section 21–12 for the current as a function of time in an LR circuit.
$$I=\frac{V}{R}(1-e^{-tR/L})=I_{maximum}(1-e^{-tR/L})$$
We are told that the maximum current is 55 mA, and after 120 microseconds, the current should be 7.5 mA. Solve for the required inductance.
$$ e^{-tR/L}=1-\frac{I}{I_{maximum}}$$
$$ L=-\frac{tR}{ln(1-\frac{I}{I_{maximum}})}$$
$$ L=-\frac{(0.00012s)(120\Omega)}{ln(1-\frac{7.5mA}{55mA})}\approx 98mH$$
Put a 98 mH inductance in series with the device.
