V2 | V3 | Ibat.1 | Ibat.2 | Pbat1 | Pbat2 |
10.26 V | 8.67 V | 17.4 mA | 1.50 mA | 0.209 W | 0.0135 W |
Next, we measured the actual values for the resistors, and set everything up on a breadboard. Here is our circuit:
We measured the actual values for the currents through source 1 and source 2 and voltage across the two loads:
Variable | Theroetical | Measured | % Error |
I bat 1 | 17.4 mA | 17.9 mA | 2.87% |
I bat 2 | 1.50 mA | 1.47 mA | -2.00% |
V2 | 10.26 V | 10.35 V | 0.88% |
V3 | 8.67 V | 8.75 V | 0.92% |
Then, we calculated power delivered by the batteries:
P bat 1 = 217 mW
P bat 2 = 13.3 mW
For the last step, we assumed that V2 and V3 were 9V, and used our previous equations to solve for the corresponding source voltages. We calculated that Vbat1 should be 9.9V and Vbat2 should be 10.98V. To achieve this source voltage drop, we hooked up resistor boxes in series with the two sources and tweaked the resistances until the desired 9V was achieved across the loads. These resistances ended up being R1s = 203 ohms and R2s = 103 ohms.
These were the resulting values for this scenario:
V2 | V3 | Ibat.1 | Ibat.2 |
9.09V | 8.99V | 9.70 mA | 8.61 mA |
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