The current through a 25-mH inductor is 10e-t/2 A. Find the voltage at t = 3 s. Enter your answer with the correct sign and in units of mV.
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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.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?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.
- An AC circuit contains a 24 resistor, a 15.9-mH inductor, and a 13.3F capacitor connected in parallel. The circuit is connected to a 240-V, 400-Hz power supply. Find the following values. XL=XC=IR=AIL=AIC=AP=WVARsL=VARsC=IT=AVA=PF=%=Three 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?4. The current through a 10-mH inductor is 10e“²A. Find the voltage and the power at t = 3s. Ans.-11.16 mV and -24.89mW
- 1. A 2F capacitor is connected in series with a diode and a 60 resistor. The voltage of the battery is 9V. The forward voltage drop of the diode is 0.7V. Flip to close the switch at t=0. Determine the voltage and current of the capacitor at t=0, t=12s, t=24s, and t=36s. (The following table may help you with this question. Feel free to use this table (or not).) Vc VR I R1 60 H Time=0 Time=12s Time=24s Time=36sAn inductor and a resistor are in series with a sinewave voltage source. The frequency is set so that the inductive reactance is equal to the resistance. If the frequency is increased, then options: Voltage across the inductor= Source Voltage Voltage across the inductor> Voltage across the resistor Voltage across the inductor= Voltage across the resistor Voltage across the resistor > Voltage across the inductorAn alternating voltage is applied to a series circuit consisting of a resistor and iron-cored inductor and a capacitor. The current in the circuit is 0.5 A and the voltages measured are 30 V across the resistor, 48 V across the inductor, 60 V across the resistor and inductor and 90 V across the capacitor. Find (a) the combined copper and iron losses in the inductor (b) the applied voltage.
- In the circuit below, Find the voltage across the inductor at time t = 0.000909 s. The values are V1 = 146V, R = 4.4 ohms, L =0.02 henriesAn alternating voltage is applied to a series circuit consisting of a resistor and iron-cored inductor anda capacitor. The current in the circuit is 0.5 A and the voltages measured are 30 V across the resistor,48 V across the inductor, 60 V across the resistor and inductor and 90 V across the capacitor.Find (a) the combined copper and iron losses in the inductor (b) the applied voltage Answer: [(a) 3.3 W (b) 56 V]An alternating voltage is applied to a series circuit consisting of a resistor and iron-cored inductor anda capacitor. The current in the circuit is 0.5 A and the voltages measured are 30 V across the resistor,48 V across the inductor, 60 V across the resistor and inductor and 90 V across the capacitor.Find (a) the combined copper and iron losses in the inductor (b) the applied voltage.Answer: [(a) 3.3 W (b) 56 V] Note: Please provide a clear solution!