In this lab, we are modeling a redundant power system that has two voltage sources, in case one fails. This system should have breakers (in order to isolate bad components), which we are modeling with 100 ohm and 220 ohm resistors. Finally, we model two loads, each with 1K ohm resistors. Before we set up the circuit, we were able to calculate the voltage across the loads using nodal analysis, solving for the resulting system of equations. Substituting back into our equations, we could calculate current through each of the power sources and the power they supply.
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:
#1: Built the circuit.
We measured the actual values for the currents through source 1 and source 2 and voltage across the two loads:
#2: Built the circuit.
#3 set up the power
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. 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 |
9.04V | 9.00V | 10.0 mA | 8.61 mA |
#4: step 4 :set both Voltage =9V
#5: Adjust power,Reading required battery voltage=10V
#6 the other one reading 2.9V
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