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F07%20314%20HW05%20all-s

# F07%20314%20HW05%20all-s - EECS 314 Fall 2007 HW 05...

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Problem 1 Solution From the given equation for energy, we see that we need to know the voltage across each capacitor to find the energy. Each of the capacitors in this problem is in parallel with a resistor, and therefore if we know the voltage across the resistors, we know the voltage across the capacitors. We also know that in a DC circuit, capacitors are open, and no current flows through them. We can therefore neglect them in calculating the voltage across the resistors, which becomes a simple voltage division problem. Redrawing the circuit with the capacitors neglected (since there is no current through them): Now solve for each of the voltages using voltage division: V V V S R 75 ) 8 2 //( ) 10 20 ( 5 . 7 5 . 7 1 V V V S R 25 ) 8 2 //( ) 10 20 ( 5 . 7 ) 8 2 //( ) 10 20 ( 20 10 10 2 V V V S R 60 ) 8 2 //( ) 10 20 ( 5 . 7 ) 8 2 //( ) 10 20 ( 2 8 8 3 Since the capacitors are in parallel with these resistors, they have the same voltage, and therefore: EECS 314 Fall 2007 HW 05 solutions Page 1 of 12

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V V V R nF 75 1 15 V V V R nF 25 2 25 V V V R nF 60 3 200 Using W 1 2 C V 2 , we get mJ V F W nF 2 . 42 ) 75 )( 15 ( 2 1 2 15 mJ V F W nF 1 . 78 ) 25 )( 250 ( 2 1 2 25 mJ V F W nF 720 ) 60 )( 400 ( 2 1 2 200 EECS 314 Fall 2007 HW 05 solutions Page 2 of 12
Problem 2 To find the energy stored in each inductor, first note that all of the inductors are in DC steady state conditions, which means we can replace all of them with short circuits: Note that the 1 Ω , 16 Ω , and 20 Ω resistors have been shorted out by the inductors, which are represented as wires under DC steady-state conditions. Because they have

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