(a) Find the total inductance, LT, in the network below. All inductors are in millihenrys. LT 3 www 8 9 18 9
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- 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?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?A capacitor in series with a 200 ohms resistor draws a current of 0.3 ampere from 120 volts, 60 hz source. What is the value of the capacitor in microfarad? a. 8.7 b. 9.7 c. 6.7 d. 7.7
- 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 -24.89mWA resistor of 20ohms and a capacitance of unknown value when connected. In in parallel across a 100V, 50Hz supply take a current of 6A. The combination is now connected across a 100V supply of unknown frequency and the current falls to 5.5A . What is the frequency?Find the equivalent inductance as seen from the terminals shown in the figure.
- At 0-, no currrent flows through the capacitors because they are open, how did you combine the capacitors for the voltage divider since it is the capacitance value and not reactance, is it right? Can we just combine the capacitance value? Please explain. I did not understand..A resistor of 20 Ohms and a capacitance of unknown value when connected in in parallel across a 100V, 50Hz supply, take a current of 6A. The combination is now connected across a 100v supply of unknown frequency and the current falls to 5.5A. What is the frequency?What impedance vector (0- j15) Ohms represents:A. A pure resistance. C. A pure capacitance.B. A pure inductance. D. An inductance combined with a capacitance.
- The current through a 50 mH inductor is given below. Use Ohm’s Law to find the voltage. Convert to polar form, find the impedance and complete the math, then convert back to the waveform at i1 = 7.5sin(2500t – 20°) mA and i2 = 5.9sin(200t + 40°) mAPls. write solution neatlyAs discussed in your reading material, the MAIN reason for calculating a capacitance value is to adjust capacitor timing determine the nominal value of the capacitor check for short circuits check for open plates in the capacitor a. b. C. d.