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W09 314 HW05 solution - W09 EECS 314 Homework 5 Solution...

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Solution by Idong Ebong W09 EECS 314 Homework 5 Solution Problem 1 Part 1 Part 2: V S =15V; C 1 =C 2 = 2mF; L 1 =L 2 =1mH; R 1 =R 3 =4Ω ; R 2 =R 5 =1Ω ; R 4 =R 6 Capacitor Voltages =2Ω V C1 = V R2 = R 2 R 1 + R 2 V S = 1 Ω 4 Ω + 1 Ω 15V = 3V V C2 = V S = 15V I L1 = I R4 = V S R 4 = 15 V 2 Ω = 7.5 A I L2 = I R6 = V S R 6 = 15 V 2 Ω = 7.5 A Inductor Currents Component C C 1 Component 2 L L 1 2 Voltage 3V 15V Current 7.5A 7.5A Part 3: E C1 = 1 2 C 1 V C1 2 = 1 2 2 × 10 3 𝐹𝐹 ∗ (3 𝑉𝑉 ) 2 = 0.009 𝐽𝐽 E C2 = 1 2 C 2 V C2 2 = 1 2 2 × 10 3 𝐹𝐹 ∗ (15 𝑉𝑉 ) 2 = 0.225 𝐽𝐽 E L1 = 1 2 L 1 I L1 2 = 1 2 1 × 10 3 𝐻𝐻 ∗ (7.5 𝐴𝐴 ) 2 = 0.0281 𝐽𝐽 E L2 = 1 2 L 2 I L2 2 = 1 2 1 × 10 3 𝐻𝐻 ∗ (7.5 𝐴𝐴 ) 2 = 0.0281 𝐽𝐽 Component C C 1 L 2 L 1 2 Energy in Joules 0.009 0.225 0.0281 0.0281
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EECS 314 Winter 2009 Homework set 5 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 2 solution Part 1: Calculate Cab in nF: Since there is no connection between node d and any other point in the circuit, no current will flow through it and thus the 50nF capacitor can be neglected in the calculation of Cab. The 60nF and 40 nF caps are in series so the effective capacitance from those two is 60*40/(60+40) = 2400/100 = 24nF. The 24nF effective capacitor is in parallel with the 20nF capacitor and thus they add. The effective capacitance of that configuration is 20nF + 24nF = 44nF.
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