Other Single-Loop Circuits — Question 2

Question 2

Reversing one battery changes an emf sum into a difference

Two ideal sources of 6.00 V and 4.00 V are initially series aiding in a loop with total resistance 5.00 ohm. The 4.00 V source is then reversed.

1. Find current before and after the reversal.

2. Find the ratio of resistor heating powers and explain why it is not equal to the current ratio.

Question 2 - Solution

Reversing one battery changes an emf sum into a difference

(a) Method and calculation

$$I_{\mathrm{aiding}}=(6+4)/5=2.00\,\mathrm A.$$$$I_{\mathrm{opposing}}=(6-4)/5=.400\,\mathrm A.$$

(b) Interpretation and further calculation

$$P_{\mathrm{aiding}}=2^2(5)=20.0\,\mathrm W,\quad P_{\mathrm{opposing}}=.4^2(5)=.800\,\mathrm W.$$$$P_{\mathrm{opposing}}/P_{\mathrm{aiding}}=.0400=(.400/2.00)^2.$$

Physical check

The drawing shows the initial aiding connection; reversing the second source changes only its signed emf.