Other Single-Loop Circuits — Question 3

Question 3

One reversed cell weakens a series battery while retaining its resistance

A battery contains N identical cells of emf e0 and internal resistance r, in series with load R. Exactly one cell is installed opposite to the other N-1 cells. Define positive current in the direction driven by the majority orientation.

1. Find current and compare it with the all-correct battery.

2. Find the ratio for N=5 and interpret the cases N=2 and N=1 algebraically.

Question 3 - Solution

One reversed cell weakens a series battery while retaining its resistance

(a) Method and calculation

$$E_{\mathrm{net}}=(N-1)e_0-e_0=(N-2)e_0,\quad R_{\mathrm{total}}=R+Nr.$$$$I_{\mathrm{reversed}}=(N-2)e_0/(R+Nr),\quad I_{\mathrm{correct}}=Ne_0/(R+Nr).$$

(b) Interpretation and further calculation

$$I_{\mathrm{reversed}}/I_{\mathrm{correct}}=(N-2)/N=3/5\quad(N=5).$$$$N=2:\ I=0;\qquad N=1:\ I=-e_0/(R+r).$$For one cell, positive direction means its intended installation direction.

Physical check

A reversed cell still adds positive internal resistance; only its emf changes sign.