Multiloop Circuits — Question 2

Question 2

Two batteries sharing a load can change from joint supply to charging

Two sources have emfs 12.0 V and 6.00 V, each with internal resistance 2.00 ohm. Their positive branches join node V, and their negative terminals share ground. A load R joins V to ground. Define I1 and I2 toward V from the sources.

1. Derive V(R), I1 and I2, then evaluate them at R=2.00 ohm.

2. Determine the load range for which the 6.00 V source absorbs chemical energy.

Question 2 - Solution

Two batteries sharing a load can change from joint supply to charging

(a) Method and calculation

$$\frac{12-V}{2}+\frac{6-V}{2}=\frac VR\Rightarrow V=\frac{9R}{R+1\,\Omega}\quad\text{(volts when R is in ohms)}.$$$$R=2.00\,\Omega:\quad V=6.00\,\mathrm V,\ I_1=3.00\,\mathrm A,\ I_2=0.$$

(b) Interpretation and further calculation

$$I_1=(12\,\mathrm V-V)/(2\,\Omega),\quad I_2=(6\,\mathrm V-V)/(2\,\Omega).$$$$I_2<0\Longleftrightarrow V>6.00\,\mathrm V\Longleftrightarrow R>2.00\,\Omega.$$

Physical check

A negative signed source current indicates charging; the internal resistor still dissipates positive power.