Multiloop Circuits — Question 3

Question 3

A shared resistor carries the difference of clockwise mesh currents

The circuit has left and right upward emfs 12.0 V and 3.00 V. Each outer branch has 3.00 ohm resistance, and the central shared resistor is 3.00 ohm. Define clockwise mesh currents i1 in the left loop and i2 in the right loop.

1. Write and solve the two mesh equations. Find the downward current in the shared resistor.

2. Find the signed powers delivered by both sources and compare their sum with resistor heating.

Question 3 - Solution

A shared resistor carries the difference of clockwise mesh currents

(a) Method and calculation

$$6i_1-3i_2=12,\quad-3i_1+6i_2=-3.$$$$i_1=7/3\,\mathrm A,\quad i_2=2/3\,\mathrm A,\quad I_c=i_1-i_2=5/3\,\mathrm A.$$

(b) Interpretation and further calculation

$$P_{12}=12i_1=28.0\,\mathrm W,\quad P_3=-3i_2=-2.00\,\mathrm W.$$$$P_R=3[(7/3)^2+(2/3)^2+(5/3)^2]=26.0\,\mathrm W=P_{12}+P_3.$$

Physical check

Both mesh currents are clockwise, but they traverse the shared branch in opposite directions.