Other Single-Loop Circuits — Question 4

Question 4

A weak aiding cell can absorb terminal power despite releasing chemical energy

Two cells are series aiding. Their emfs are 12.0 V and 1.00 V, their internal resistances .200 and 2.00 ohm, and the external load is 1.00 ohm.

1. Find loop current and terminal voltage of the weak cell, using its labeled positive minus negative terminals.

2. Separate its chemical power, internal heat and delivered terminal power; interpret a negative terminal result.

Question 4 - Solution

A weak aiding cell can absorb terminal power despite releasing chemical energy

(a) Method and calculation

$$I=(12+1)/(.2+2+1)=4.0625\,\mathrm A.$$$$V_{\mathrm{weak}}=1.00-I(2.00)=-7.125\,\mathrm V.$$

(b) Interpretation and further calculation

$$P_{\mathrm{chem,weak}}=1.00I=4.0625\,\mathrm W,\quad P_{r,\mathrm{weak}}=2I^2=33.0078\,\mathrm W.$$$$P_{\mathrm{terminal,weak}}=V_{\mathrm{weak}}I=-28.9453\,\mathrm W=P_{\mathrm{chem,weak}}-P_{r,\mathrm{weak}}.$$

Physical check

The idealized weak cell heats using both its own chemical energy and electrical energy received from the stronger cell.