Question

A Review Constants Part E - Determine the equivalent capacitance across terminals a and b Learning Goal: To reduce series-parallel combinations of inductors or capacitors to an equivalent inductance or capacitance. Inductors in series and parallel combine like resistors in series and parallel. It is possible to use Kirchhoff's current law to find the current through the equivalent inductance. Moreover, capacitors in series combine like resistors in parallel and vice versa. It is possible to use Kirchhoff's voltage law to find the voltage across the equivalent capacitance. For the network in Figure 2(Figure 2), determine the equivalent capacitance across terminals a and b. Assume that v1 = 2.0 V, 22 = 10.0 V, v3 = 10.0 V, 05 = 10.0 V, 06 = 5.0 V, C = 8.0 uF, C2 = 10.0 uF, C3 = 2.0 uF, C4 = 10.0 uF C5 = 4.0 uF, C8 = 4.0 uF, C, = 3.0 uF, and Cg = 10.0 uF Express the equivalent capacitance across terminals a and b in microfarads to three significant figures. View Available Hint(s) ΟΙ ΑΣφ Jt vec ? Cab = 16.17 Submit Previous Answers X Incorrect; Try Again; 3 attempts remaining Part F - Determine the current, i 1, through the equivalent inductance to the right of terminals a and d and d. Assume that i3 = -1.0 mA , 14 = 1.0 mA, is = 4.0 mA, L1 = 1.0 mH, L2 = 3.0 Figure < 2 of 4 For the network in Figure 3(Figure 3), determine the current, 21, through the equivalent inductance to the right of terminals mH, L3 = 2.0 mH L4 = 10.0 mH and L5 = 8.0 mH Express your answer in milliamperes to three significant figures. C C b View Available Hint(s) Vi + + V2 E AZO It vec ? Vs V6 ij = mA 17 Submit + V8 Figure 2 Part G - Determine the voltage, V 2, across the equivalent capacitance at terminals a and b

Answer 1

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Ceas and ceous are in Series - Ceas x Ceaus Ceauzt (pas crub 4.571 X 11.6 4.571+11.6 3.27% HF Total Equivalen capacitance : C₂ + Cear 6 10+ 3.278 - 13.27% HF part 2. :? I g=-1.0mA, I 4 = 1.0 mA 15 = 4.0mA KCL at node "R" Applying i = izt i4 ti5 6 -1.0+ 1.0+40 4 mA part G KVL in loop-I 6. Applying H4 5 V4 = V, + V3 сч -9-1 16 po Ju = -lov Applying kvL in loop - Il V2 -V5-V6-V4 - 0 V2: V5 + 6 + vy 6-8 tip-10 17 v