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Unformatted text preview: Use the negative resistance converter, entry ( h ) in Figure 6.61. P7.213 We’ll write and solve a node equation. Label the node voltages as shown. Apply KCL at node a to get so Then P7.31 Given The capacitor voltage is given by For In particular, For In particular, For Now the power and energy are calculated as ___________________ Actually, it’s 0.04(t2) 3 for 2<t<6 and _______________ Actually, it’s 0.64(t4) 2 for 6<t Note: graphs were not required....
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This note was uploaded on 02/16/2010 for the course ECE ECE 35 taught by Professor Josephford during the Winter '10 term at UCSD.
 Winter '10
 JosephFord

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