Multiloop Circuits — Question 6

Question 6

Voltage superposition does not imply power superposition

The two-source network has E1=12.0 V, r1=3.00 ohm, E2=6.00 V, r2=6.00 ohm and load R=2.00 ohm. To isolate one source contribution, set the other ideal emf to zero and keep its resistor.

1. Find the load voltages V1 and V2 due to each source and their combined value.

2. Compare actual load power with the sum of powers calculated from V1 and V2 separately; identify the missing term.

Question 6 - Solution

Voltage superposition does not imply power superposition

(a) Method and calculation

$$G=1/3+1/6+1/2=1.00\,\mathrm S,\quad V_1=(12/3)/G=4.00\,\mathrm V.$$$$V_2=(6/6)/G=1.00\,\mathrm V,\quad V=V_1+V_2=5.00\,\mathrm V.$$

(b) Interpretation and further calculation

$$P=\frac{(V_1+V_2)^2}{R}=12.5\,\mathrm W,\quad P_1+P_2=\frac{V_1^2+V_2^2}{R}=8.50\,\mathrm W.$$$$P-P_1-P_2=\frac{2V_1V_2}{R}=4.00\,\mathrm W.$$

Physical check

Linear circuit variables superpose; quadratic energy and power generally do not.