Work, Energy, and Emf — Question 4

Question 4

A declining emf requires an energy integral over delivered charge

A source emf depends on cumulative delivered charge q as $\mathcal E(q)=E_0(1-q/(2Q_0))$ for $0\le q\le Q_0$. It discharges at constant current I through internal resistance r, with $Ir<E_0/2$.

1. Find chemical energy released and terminal energy delivered over the full interval.

2. Find average terminal voltage and overall energy-delivery efficiency.

Question 4 - Solution

A declining emf requires an energy integral over delivered charge

(a) Method and calculation

$$W_{\mathrm{chem}}=\int_0^{Q_0}\mathcal E(q)dq=\tfrac34E_0Q_0.$$$$W_{\mathrm{out}}=\int_0^{Q_0}[\mathcal E(q)-Ir]dq=(\tfrac34E_0-Ir)Q_0.$$

(b) Interpretation and further calculation

$$\overline V=W_{\mathrm{out}}/Q_0=\tfrac34E_0-Ir.$$$$\eta=W_{\mathrm{out}}/W_{\mathrm{chem}}=1-\frac{4Ir}{3E_0}.$$

Physical check

Emf is energy per incremental charge; multiplying only the initial emf by total charge overestimates released energy.