 # 3.Solutions - DC Circuits (1).pdf - Part A – Series and...

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Part A – Series and Parallel resistorsThe actual voltage that you choose for your battery is irrelevant, you may end up with slightly different rounded valuesthan what I did though if you have something different.Problem 1:Experimental𝑅=𝑉𝐼=1000.57= 175.4ΩTheoretical𝑅𝑒𝑞=𝑅1+𝑅2+𝑅3= 25 + 50 + 100= 175ΩCan chalk up the difference to rounding error.Problem 2:Experimental𝑅=𝑉𝐼=1007= 14.3ΩTheoretical1𝑅𝑒𝑞=1𝑅1+1𝑅2+1𝑅3=125+150+1100=7100𝑅𝑒𝑞=1007= 14.3ΩIn this case it is an exact matchProblem 3:Experimental𝑅=𝑉𝐼=1001.71= 58.5ΩTheoreticalGoing to do this in batches, first the 50 & 100Ωresistorsin parallel
1𝑅23=1𝑅2+1𝑅3=150+1100=3100∴ 𝑅23=1003= 33.3ΩNow the equivalent circuit is𝑅𝑒𝑞=𝑅1+𝑅23= 25 + 33.3 = 58.3ΩClose enough to call it a rounding errorProblem 4:Experimental𝑅=𝑉𝐼=1000.86= 116.3ΩTheoreticalGoing to do this in batches, first the 25 & 50Ωresistors in parallel1𝑅12=1𝑅1+1𝑅2=125+150=350∴ 𝑅12=503= 16.7ΩNow the equivalent circuit is𝑅𝑒𝑞=𝑅12+𝑅3= 16.7 + 100 = 116.7ΩClose enough to call it a rounding errorProblem 5:Experimental𝑅=𝑉𝐼=1002.33= 42.9ΩTheoreticalGoing to do this in batches, first the 25 & 50Ωresistors in series𝑅12=𝑅1+𝑅2= 25 + 50 = 75Ω
1𝑅𝑒𝑞=1𝑅12+1𝑅3=175+1100=7300𝑅𝑒𝑞=3007= 42.9ΩLooks good to me!Part B: The BatteryOhm’s law gets the current based off thevoltageRx ()Vx (V)Ix(A)5.21.9690.3786546.12.2090.3621318.12.6280.32444492.7890.30988910.93.0780.282385123.2110.26758313.93.4270.24654714.93.5310.2369819.83.9130.19762629.94.40.14715749.64.880.098387So we see the slope of the graph is -10.228 V/A = -10.228 ΩDropping the negative sign, I’ll explain why we can get away with this later and in the discussions, we see thatthe totalinternal resistance is 10.228Ωwhich is both the resistance of the battery and the limiting resistor.

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Term
Fall
Professor
Alfred
Tags
Resistor, Electrical resistance, Vint, Series and parallel circuits, 0 58 amps, 5 93 volts
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