C1=5 µF, C2=10 µF, C3=9 µF and C4=14 uF C1 С4 C3 C2 B The value of equivalent capacitance (in µF) between A and B
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- You are an electrician working in an industrial plant. You discover that the problem with a certain machine is a defective capacitor. The capacitor is connected to a 240-volt AC circuit. The information on the capacitor reveals that it has a capacitance value of 10 mF and a voltage rating of 240 VAC. The only 10-mF AC capacitor in the storeroom is marked with a voltage rating of 350 WVDC. Can this capacitor be used to replace the defective capacitor? Explain your answer.You find that a 25-F capacitor connected to 480 VAC is defective. The storeroom has no capacitors with a 480-VAC rating. However, you find two capacitors rated at 50 F and 370 VAC. Can these two capacitors be connected in such a manner that they can replace the defective capacitor? If yes, explain how they are connected and why the capacitors will not be damaged by the lower voltage rating. If no, explain why they cannot be used without damaging the capacitor.A capacitor uses air as a dielectric and has a capacitance of 3 F. A dielectric material is inserted between the plates without changing the spacing, and the capacitance becomes 15 F. What is the dielectric constant of this material?
- A 15-F AC capacitor is connected in series with a 50 resistor. The capacitor has a voltage rating of 600 WVDC. The capacitor and resistor are connected to a 480-V, 60-Hz circuit. Is the voltage rating of the capacitor sufficient for this connection?Inductive Circuits Fill in all the missing values. Refer to the following formulas: XL=2fLL=XL2ff=XL2L Inductance (H) Frequency (Hz) Inductive Reactance ( ) 1.2 60 0.085 213.628 1000 4712.389 0.65 600 3.6 678.584 25 411.459 0.5 60 0.85 6408.849 20 201.062 0.45 400 4.8 2412.743 1000 40.841Three capacitors having capacitance values of 20F,40F, and 50F are connected in parallel to a 60 - Hz power line. An ammeter indicates a circuit current of 8.6 amperes. How much current is flowing through the 40F capacitor?
- A postage stamp mica capacitor has the following color marks starting at the upper left dot: yellow, violet, brown, green, no color, and blue. What are the capacitance value, tolerance, and voltage rating of this capacitor?You are working in an industrial plant. You have been instructed to double the capacitance connected to a machine. The markings on the capacitor, however, are not visible. The capacitor is connected to 560 volts and an ammeter indicates a current of 6 amperes flowing to the capacitor. What size capacitor should be connected in parallel with the existing capacitor? What is the minimum AC voltage rating of the new capacitor? What is the minimum DC voltage rating of the new capacitor? What is the minimum KVAR size that can be used in this installation?C3 Vin \aov AC Cr D3 Vin Voat /calculate Vout and explain the followingcircait. ALl capacitor 0,4MF And Calculate the energy trapped in each capacitor and the total energy trapped in the circuit ell
- 2 A C₁ Cz C3 C4 B 11-1 Cf C6 C₁=C₂ = C3 = 2μF 4mF C4 = Cr²C6 TOTAL CAPACITANCE FROM AOINT AB? TOTAL ENERGY STORED?Calculate values of asked physical quantities of the circuit in the figure. What is the capacitance of system 23? C23 C₁ = 3.0 F What is the capacitance of system 123? C123 = F C₁ = 2.0 F What is the capacitance of system 45? C45 = F Your last answer was interpreted as follows: 1 Your last answer was interpreted as follows: 1 What is the charge in the capacitor system? C Your last answer was interpreted as follows: 1 What is the total capacitance of the system? F Value 8.8500 x 10 EO Quantity permittivity of free space capacitance capacitance capacitance capacitance capacitance Your last answer was interpreted as follows: 1 capacitance of system 23 capacitance of system 123 capacitance of system 45 Your last answer was interpreted as follows: 1 Do not leave any fields blank. If you don't know the answer, insert 1 for example. Insert the answer with 3 significant digits without rounding the answer. Constant U = 7.0 V Unit E Symbol €0 C₁ = 3.0 F C1 C₂ C3 C4 C5 C23 C₂ = 9.0 F C₂ = 3.0 F…find total capacitance and total charge. C1 = 20pF C2 = 30pF 5V C3 = 30pF C5 10pF C4 30pF