Calculating the Current in a Single-Loop Circuit — Question 2

Question 2

An inconvenient current arrow is corrected by the algebraic sign

An ideal source of emf E has its positive terminal at the top of the drawn loop. A single resistor R occupies the top branch. Define trial current J counterclockwise, opposite the direction suggested by the source.

1. Write the counterclockwise loop equation and solve for J.

2. Express the actual current and resistor power without changing the original definition of J.

Question 2 - Solution

An inconvenient current arrow is corrected by the algebraic sign

(a) Method and calculation

$$-E-JR=0.$$$$J=-E/R<0.$$

(b) Interpretation and further calculation

$$I_{\mathrm{clockwise}}=-J=E/R.$$$$P_R=J^2R=E^2/R>0.$$The negative result reverses the trial arrow, not the resistor sign convention.

Physical check

A consistent sign convention works even when the current direction is initially guessed incorrectly.