Question

The National Ignition Facility at the Lawrence Livermore National Laboratory maintains the world’s largest laser facility. The lasers, which are used to start a nuclear fusion reaction, are powered by a capacitor bank that stores a total of about 400 megajoules of energy. When the lasers are fired the capacitors discharge completely and then immediately begin recharging. The charge Q of the capacitors \(t\) seconds after the discharge is given by

\(Q\left( t \right) = {Q_0}\left( {1 - {e^{ - \frac{t}{a}}}} \right)\)

(The maximum charge capacity is \({Q_0}\) and \(t\) is measured in seconds.)

(a) Find a formula for the inverse of this function and explain its meaning.

(b) How long does it take to recharge the capacitors to \(90\% \) of capacity if \(a = 50\)?

Short answer

(a) The inverse function is \({f^{ - 1}}\left( Q \right) = - a\ln \left( {1 - \frac{Q}{{{Q_0}}}} \right)\). The formula provides the time needed to get the given charge \(Q\) after a discharge.

(b) The capacitors will take approximately 115 seconds to recharge to 90% of capacity.

Step-by-step solution

Determine the formula for the inverse of the function

a)

Write the equation as \(Q = {Q_0}\left( {1 - {e^{ - \frac{t}{a}}}} \right)\) and solve the equation for \(t\) as shown below:

\(\begin{aligned}Q &= {Q_0}\left( {1 - {e^{ - \frac{t}{a}}}} \right)\\\frac{Q}{{{Q_0}}} &= 1 - {e^{ - \frac{t}{a}}}\\{e^{ - \frac{t}{a}}} &= 1 - \frac{Q}{{{Q_0}}}\\ - \frac{t}{a} &= \ln \left( {1 - \frac{Q}{{{Q_0}}}} \right)\\t &= - a\ln \left( {1 - \frac{Q}{{{Q_0}}}} \right)\end{aligned}\)

Therefore, the inverse function is \({f^{ - 1}}\left( Q \right) = - a\ln \left( {1 - \frac{Q}{{{Q_0}}}} \right)\).

The formula shows the time (in seconds) needed to obtain a given charge \(Q\) after a discharge.

Determine the time it takes to recharge the capacitor to 90%

b)

Let \(Q = 0.9{Q_0}\) and \(a = 50\).

The time it takes to recharge the capacitor to 90% as shown below:

\(\begin{aligned}t &= - 50\ln \left( {1 - \frac{{0.9{Q_0}}}{{{Q_0}}}} \right)\\ &= - 50 \cdot \ln \left( {0.1} \right)\\ &= - 50 \cdot - 2.302\\ \approx 115.12\end{aligned}\)

Thus, the capacitors will be recharged to 90% of capacity in approximately 115 seconds short of two minutes.