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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?C5 What is the correct explanation for the voltage seen at Va in Figure C5? Va 1 Sost We will assume that the Vout Out Vb, capacitor is charged to a value determined by the feedback and the supply voltage +BVsat. Figure C5 A. The waveform on the capacitor looks like a discharge, and then a charging waveform to an equal but opposite magnitude voltage, B. The capacitor simply reverses its charge, with a charging waveform, C. The capacitor simply reverses its charge, with a discharging waveform, D. None of the above.Design a circuit to limit a 20 V rms sinusoidal voltage to a maximum positive amplitude of 10 V and a maximum negative amplitude of using a single 14 V dc voltage source.
- Example 4: Determine the Q waveform and compare it with the input waveform. Assume Q = 0 initially. Cik Cik Data Input Solution) Data Inpur DDesign a circuit to limit a 20 V rms sinusoidal voltage to a maximum positive amplitude of 10 V and a maximum negative amplitude of -5 V using a single 14 V de voltage source.Problem Design a circuit to limit a 20 V rms sinusoidal voltage to a maximum positive amplitude of 10 V and a maximum negative amplitude of -5 V using a single 14 V dc voltage source.
- A pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) A (b) 100 - A TC (c) 100 A (d)70.7 Aplease solve value of Capacitor with 0.1V Ripple. And draw output ripple waveform. Explain it one by one for me thanks..One steradian represents .... .... ... (rad)^2 O degree O 2rad (rad)3 O
- Q1(a) Analyze the polar form of this waveform in Figure Q1(a). 60 V 310⁰ Figure Q1(a)A pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) 100 (b) - - (c) 100 A (d) 70.7 AShown in the figure below is an "RL" circuit drive by an AC power source. The AC power source has an RMS voltage of Vps (RMS) = 9.84 Volts and is running at a frequency of f = 8.585e+04 Hz. The resistor has a resistance of R = 2170 and the inductor has an inductance of L = 3.54e-03 Henries. Vps R ww Write the FORMULA for the total impedance of the circuit Ztot = Determine the numerical value of Ztot = 2890.5 Determine the numerical value of $z= = 41 Determine the current through the circuit: • I(PEAK) = 4.81E-3 • I(RMS) = 3.404E-3 Determine the voltage across the resistor: • VR(PEAK) = 7.387 • VR(RMS) = 5.22 ✔✔ Amps ✔Amps Write the FORMULA for the phase of the total impedance of the circuit z... = | tan-1 2701 R x Volts X Volts Determine the voltage across the inductor: • VL(PEAK) = 9.184 • VL(RMS) = 6.49 ✔ Volts Volts L R²+ WL- ✔ degrees 2 If a second circuit were connected in parallel with the inductor, this circuit would be considered as: O a low-pass filter O a capacitive switcher…