# hmwk 5 sol - EECS 314 Fall 2006 HW 05 Problem 1 Student's...

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EECS 314 Fall 2006 HW 05 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) The big picture 1. Capacitors store energy in the form of electric field. The energy W stored in the capacitor equals W = 1 2 C V 2 where C is the capacitance and V is the voltage across the capacitor. Units: if C is in farads and V is in volts, then W is in joules. 2. Capacitors act as open circuits under DC steady-state conditions. They do not conduct DC current, and the voltage across the capacitor is determined by the voltage across the circuit elements (sources, resistors, etc.) with which this capacitor is in parallel. Problem Calculate the energy stored in each capacitor in this circuit. Assume DC steady- state conditions. Show your work. Problem 1 Solution Page 1 of 3

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EECS 314 Fall 2006 HW 05 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) Solution (Barry Mullins) 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 8 . 28 ) 2 10 //( ) 8 40 ( 4 . 14 4 . 14 1 = + + + = V V V S R 16 ) 2 10 //( ) 8 40 ( 4 . 14 ) 2 10 //( ) 8 40 ( 8 40 40 2 = + + + + + + = V V V S R 2 . 3 ) 2 10 //( ) 8 40 ( 4 . 14 ) 2 10 //( ) 8 40 ( 2 10 2 3 = + + + + + + = Since the capacitors are in parallel with these resistors, they have the same voltage, and therefore: Problem 1 Solution Page 2 of 3
EECS 314 Fall 2006 HW 05 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) V V V R nF 8 . 28 1 15 = = V V V R nF 16 2 25 = = V V V R nF 2 . 3 3 200 = = Using W = 1 2 C V 2 , we get uJ V nF W nF 22 . 6 ) 8 . 28 )( 15 ( 2 1 2 15 = = uJ V nF W nF 2 . 3 ) 16 )( 25 ( 2 1 2 25 = = uJ V nF W nF 024 . 1 ) 2 . 3 )( 200 ( 2 1 2 200 = = Problem 1 Solution Page 3 of 3

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EECS 314 Fall 2006 HW 05 Problem 2 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) The big picture 1. Inductors store energy in the form of magnetic field. The energy W stored in the inductor equals W = 1 2 L I 2 where L is the inductance and I is current through the inductor. Units: if L is in henrys and I is in amps, then W is in joules. 2.
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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 5 sol - EECS 314 Fall 2006 HW 05 Problem 1 Student's...

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