Superposition: in a linear circuit, each voltage or current is the sum of the values caused by each independent source acting alone.
- V1 alone (V2 shorted): R2 parallel R3 = 2 ohm, so V_A' = 10 x 2/4 = 5 V
- V2 alone (V1 shorted): R1 parallel R2 = 1.333 ohm, so V_A'' = 6 x 1.333/5.333 = 1.5 V
- Total: 6.5 V, the nodal answer
Power does not superpose. The partial powers in R2 add to 6.25 + 0.5625 = 6.8125 W, but the true value is 6.5 squared / 4 = 10.5625 W. Add voltages or currents, then compute power once.
Maximum power transfer: a load draws most power when its resistance equals R_TH, and P_max = V_TH squared / (4 R_TH).
| Load on the 6.667 V, 1.333 ohm source | Load power |
|---|
| 4 ohm | 1.25 squared x 4 = 6.25 W |
| 1.333 ohm (matched) | 44.44 / 5.333 = 8.333 W |
Efficiency when matched is only 50 per cent, so power systems avoid it while signal circuits use it. In AC, the load should be the complex conjugate of the source impedance.