Question 4 For the network shown below, the voltage across the capacitor $C_2$ is: 60 V A. 38.18 V B. 25.66V C. 51.25 V D. 42.15 V $C_2$ 40 µF $C_1$ 20 µF $C_3$ 70 µF
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We can do this by adding the individual capacitances in parallel: 1/C_total = 1/C2 + 1/C3 1/C_total = 1/40 + 1/70 1/C_total = (7 + 4)/280 1/C_total = 11/280 C_total = 280/11 Show more…
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