we used the DC circuit to calculate the power by using P=IR^2 from the cable and resistor, and then using formula P=IV to get the power from power supply.
1. load is rated to consume 0.144w and supplied 12v, by R=V^2/p=144/0.144= 1000ohm
The resistor color of 1000 ohm is brown black Red.
Then using the multimeter to measure the resistor and get the measured value is 981ohm
2. using the multimeter to get the measured value of power supply, V= 12.18volt
3. the resistor box is 78 ohm
The data get from the step 6 perform the experiment.
using the multimeter connet to the circult.
Vload = 11.24volt
Ibatt= 11.4mA
A)
B)
Power to the load =Vload*I= 0.128136 watt
power to the cable= I^2*R= 0.010137 watt
the efficiency= Pout/(Pout +Plost)=Pload/(Pload+ Pcable)= 92.66%
C)
we didn't exceeding the power capability of resistor box. for I=V/(R+Rbox)
Pbox=I^2=V^2/(Rbox*(R+Rbox)^2)
The power of box will be bigger when Rbox very small.
D)
78ohm/ (0.3451ohm/m)[AEG#30] =226.021m (all cable)
E)
F)
The volt down (48-36)=12volt, use R=V/I=12/10=1.2ohm
the resistance of cable is 1.2ohm
| AWG | Resistance(ohm/ft) | Cable maximun length |
|---|---|---|
| 10 | .00118 | 1016.9 |
| 12 | .00187 | 641.7 |
| 14 | .00297 | 404.04 |
| 16 | .00473 | 253.7 |
| 18 | .00751 | 159.8 |
| 20 | .0119 | 100.8 |
| 22 | .0190 | 63.2 |
| 24 | .0302 | 39.7 |
| 26 | .0480 | 25 |
| 28 | .0764 | 15.7 |
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