Energy Stored in an Electric Field — Question 10

Question 10

A finite voltage window leaves part of the capacitor energy unavailable

A storage capacitor must deliver useful energy W=2.00 J while its voltage falls from 100 V to 60.0 V. A converter transfers fraction $\eta=.800$ of the removed capacitor energy to the load.

1. Find the minimum capacitance in this idealized energy model.

2. Find the initial stored energy, final stored energy and dissipated energy.

Question 10 - Solution

A finite voltage window leaves part of the capacitor energy unavailable

(a) Method and calculation

$$W=\eta\frac C2(V_i^2-V_f^2)\Rightarrow C_{\min}=\frac{2W}{\eta(V_i^2-V_f^2)}.$$$$C_{\min}=\frac4{.800(10000-3600)}=.00078125\,\mathrm F=781.25\,\mu\mathrm F.$$

(b) Interpretation and further calculation

$$U_i=3.90625\,\mathrm J,\quad U_f=1.40625\,\mathrm J,\quad\Delta U_{\mathrm{removed}}=2.50000\,\mathrm J.$$$$W_{\mathrm{loss}}=(1-\eta)\Delta U_{\mathrm{removed}}=.50000\,\mathrm J.$$

Physical check

Only the energy difference across the permitted voltage window is usable; the terminal 60 V state still contains stored energy.