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Three capacitors of 2 μ F, 5 μ F and 10 μ F have breakdown voltage of 200 V, 500 V and 100 V respectively. The capacitors are connected in series and the applied direct voltage to the circuit is gradually increased. Which capacitor will breakdown first ? Determine the total applied voltage and total energy stored at the point of breakdown. |
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Answer» C1 of 2μF, C2 of 5μ F, and C3 of 10μF are connected in series. If the equivalent single capacitor is C, 1/C = 1/C1 + 1/C2 + 1/C3, which gives C = 1.25 μ F If V is the applied voltage, V1 = V × C/C1 = V × (1.25/2) = 62.5 % of V V2 = V × (C/C2) = C × (1.25/5) = 25 % of V V3 = V × (C/C3) = V × (1.25/10) = 12.5 % of V If V1= 200 volts, V = 320 volts and V2 = 80 volts, V3 = 40 volts. It means that, first capacitor C1 will breakdown first. Energy stored = 1/2 CV2 = 1/2 × 1.25 × 10− 6 × 320 × 320 = 0.064 Joule |
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