R1=1kiloohm Current1= 5.05mA V1=5.06 R2=2kiloohm Current 2= 2.25mA V2=5.06 R3=5k
ID: 1308745 • Letter: R
Question
R1=1kiloohm Current1= 5.05mA V1=5.06
R2=2kiloohm Current 2= 2.25mA V2=5.06
R3=5kiloohm Current 3= 1.01mA V3=5.06
Current total = 8.60mA
Vtotal = 5.05
Lab results for a parllel circuit are above here is my questions...
1.)In a parallel circuit the sum of the circuits through each resistor should be equal to the total current supplied by the battery or power supply. Verify this property of parallel circuits by using your measured data and calculate the percentage of error.
R1=1kiloohm Current1= .62mA V1=.618
R2=2kiloohm Current 2= .62mA V2=1.24
R3=5kiloohm Current 3= .62mA V3=3.098
Current total = .62
Vtotal = 4.96
Above is a series circuit and here is my question for it.
1.)The power (VI) that is supplied by the electromotance must equal the sum of the power losses (I^2R) in each of the resistors. Using your data for the series circuit, calculate the power supplied by the battery or power supply and the power lost by each of the three resistors. If the power supplied by the source and the power lost through the three resistors are not identical, calculate the percentage error.
Explanation / Answer
for first one
Ir1+ir2+ir3=8.31 mA
which is not equal to the total current=8.6 mA
error=8.6-8.31=0.29 mA
percentage error=0.29/8.6 *100=3.37%
for second one
total power supplide=Ptotal=total current*Vtotal=0.62*4.96=3.075 mW
powe lost by R1=P1=0.62*0.618=0.38316 mW
P2=0.62*1.24=0.7688 mW
P3=0.62*3.098=1.9207 mW
P1+P2+P3=3.072mW
percentage error=
it ios not equal to 3.075 mW
so error=3.075 mW-3.072 mW=0.003
tpercentage error=0.003/3.075 *100 =0.9225 %
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