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Unformatted text preview: 7.4.2: A Parallel Circuit
R1
R3
R total
V1
V3
I1
I3
I total
Voltage Potential 104.6 Ω
318.8 Ω
80.5 Ω
6.14 V
6.14 V
56 mA
21.5 mA
78 mA
6.40V 1.
1/R = 1/ R1 + 1/ R3 = 1/104.6 Ω +1/318.8 Ω
Therefore, R = 78.8 Ω
V=IR
I’ = V/R = 6.40 V/78.8 Ω = 81.2 mA
I1’ = V/R1 = 6.40 V/104.6 Ω = 61.2 mA
I3’ = V/R3 = 6.40 V/318.8 Ω = 20.1 mA
V1’= I1’*R1 = 61.2 mA*104.6 Ω = 6.40 V
V3’= I3’*R3 = 20.1 mA*318.8 Ω = 6.40 V
2.
( /I’)*100% = ( mA81.2 mA/81.2 mA)*100% = 3.94 %
II’
78
( 1I1’/I1’)*100% = ( mA61.2 mA/61.2 mA)*100% = 8.49 %
I
56
( 3I3’/I3’)*100% = (
I
21.5 mA20.1 mA/20.1 mA)*100% = 6.97 %
The values are precise and accurate. All the measured voltages are little bit smaller values comparing to the
expected values.
3.
( 1 V/ V)*100% = (
V
6.14V6.40V/6.40V)*100% = 4.06 %
( 3 V/ V)*100% = (
V
6.14V6.40V/6.40V)*100% = 4.06 %
Both of measured values of voltage were exactly equal to each other therefore they have same percentage
differences. Because it is a parallel circuit, equal amount of voltage are sent to resistors regardless of values
of resistors.
4.
( 1+I3)  I/I)*100% = ( mA+21.5 mA) – 78 mA/78 mA)*100% = 0.64 %
(I
(56
Because it is a parallel circuit, total amount of measured current values should be equal to the sum of the
each measured current values.
5.
( – (1/ R1 + 1/ R3)/ (1/ R1 + 1/ R3))*100% = (
1/R
1/80.5 Ω – (1/ 104.6 Ω + 1/ 318.8 Ω )/ (1/ 104.6 Ω +
1/ 318.8 Ω ))*100% = 2.16 %
The values are precise and accurate. ...
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 Spring '11
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