Lab 7 data 7.4.2

Lab 7 data 7.4.2 - 7.4.2: A Parallel Circuit R1 R3 R total...

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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% = ( mA-81.2 mA/81.2 mA)*100% = 3.94 % I-I’ 78 ( 1-I1’/I1’)*100% = ( mA-61.2 mA/61.2 mA)*100% = 8.49 % I 56 ( 3-I3’/I3’)*100% = ( I 21.5 mA-20.1 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 6.14V-6.40V/6.40V)*100% = 4.06 % ( 3- V/ V)*100% = ( V 6.14V-6.40V/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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