hmwk 2 sol - Kevin Klein Homework 2 Problem 1 Solution The...

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Kevin Klein Homework 2 Problem 1 Solution The first step of this problem is to find the current through node A and the voltage at node B in the original circuit: mA I 50 . 0 7500 1500 3000 6 ! " " ! V R I V 75 . 3 7500 * 10 * 5 . 0 * 3 ! ! ! # Next, find the same values using the good instruments: mA I 5012 . 0 0 . 7472 1 . 0 1500 3000 6 ! " " " ! The 0.1 ! resistance appearing in the denominator of the current equation is the resistance of the good ammeter. The 7.4720 k ! resistance is the parallel combination of the 2 M ! voltmeter resistance and the 7.5 k ! resistor: $ ! " ! 0 . 7472 7500 10 * 2 7500 * 10 * 2 6 6 eq R
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Kevin Klein Next, find the voltage at node B: V R I V 745 . 3 7472 * 10 * 5012 . 0 * 3 ! ! ! # Next, find the same values using the so-so instruments: mA I 5091 . 0 5 6 . 7281 1500 3000 6 ! " " " ! Here, the voltmeter resistance is calculated as follows: $ ! " ! 6 . 7281 7500 10 * 250 7500 * 10 * 250 3 3 eq R Next, find the voltage at node B: V R I V 707 . 3 6 . 7281 * 10 * 5091 . 0 * 3 ! ! ! # Now the errors can be computed: Good instruments % 24 . 0 100 * 10 * 5 . 0 10 * 5 . 0 10 * 5012 . 0 3 3 3 ! # ! # # # error I % 1333 . 0 100 * 75 . 3 75 . 3 745 . 3 # ! # ! error V So-so instruments % 82 . 1 100 * 10 * 5 . 0 10 * 5 . 0 10 * 5091 . 0 3 3 3 ! # ! # # # error I % 147 . 1 100 * 75 . 3 75 . 3 707 . 3 # ! # ! error V
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Kevin Klein Now the table can be completed: Current, mA % error for current Voltage, V % error for voltage Original circuit 0.5 Zero 3.75 Zero Circuit with good instruments 0.5012 0.24 3.745 -0.1333 Circuit with so-so instruments 0.5091 1.82 3.707 -1.147
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Homework 2 Problem 2 Solution The first step is to find the resistance of the 40, 60, and 100 W lamps: R 2 24 40 ! R 2 24 60 ! R 2 24 100 ! Solving these equations gives the following resistances: " ! 4 . 14 40 R " ! 6 . 9 60 R " ! 76 . 5 100 R In circuit 2, the voltage drop across each resistor will be 20 V because they are all in parallel. Because the resistance of each lamp is known, the currents can be calculated: A I 389 . 1 4 . 14 20 40 ! ! A I 083 . 2 6 . 9 20 60 ! ! A I 472 . 3 76 . 5 20 100 ! ! The power dissipated in each lamp can now be calculated: W R I P 78 . 27 40 2 40 40 ! ! W R I P 65 . 41 60 2 60 60 ! ! W R I P 44 . 69 100 2 100 100 ! ! The power supplied by the source must equal the total power dissipated in the three lamps: # $ # $ W V I I I P S source 88 . 138 20 * 472 . 3 083 . 2 389 . 1 * 100 60 40 ! % % ! % % ! Here, the current through the source is just the sum of the currents through the lamps.
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This note was uploaded on 09/23/2009 for the course EECS EECS314 taught by Professor Ganago during the Fall '09 term at University of Michigan.

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hmwk 2 sol - Kevin Klein Homework 2 Problem 1 Solution The...

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