Q-2) Try to obtain the ratio between the average and RMS value of the signal wave given below: v(t) (V) 0 1 -9 2 4 6 t(ms) Figure 2. Circuit Schema for Problem 2.
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- For the current waveform that shown in Figure below Find:- 1). Average value 2). R.M.S value 3). Form factor Current (A) ha 8 10 12 1-2Q.2) Determine all values of resistors and capacitor for the triangular waveform generator whose output alternates between +10Vpeak and -10V peak when the input is a 10-Hz square wave that alternates between +2V and -2V. The input resistance to the generator must be at least 15 kQ. you may assume that there is no de component or offset in the input.1. Use the Gram-Schmit procedure to obtain an orthogonal basis for the set of waveforms shown in Figure 1. 0.4T T. 0.4T 0.8T 0.6T
- For the below sinusoidal voltage trace displayed by the CRO, the time/cm switch is on 5ms/cm and the volt/cm switch is on 3 V/cm. a) the frequency will bel Hz b) the amplitude will be V.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 AHow do I find the phase of this sinc function?
- The pictuere shows double-beam c.r.o. waveform traces. Find the ff:• Amplitude of P• Peak-to-peak value of Q• Periodic time of both waveforms• Frequency of both waveforms• R.m.s. value of P• R.m.s. value of Q• Phase displacement of waveform Q relativeto waveform PA certain impedance is given by 1012° N. Does the current lead or lag the voltage, and by how many degrees? Fill in the blanks. The current the voltage byA sinusoidal waveform is displayed on a C.R.O. screen, the time/cm switch is on 30 µs/cm and the volt/cm switch is on 20 V/cm. The width of one complete cycle is 3.8 cm and the peak to peak distance is 4.8 cm. Determine the following: The frequency is? The magnitude of the sine voltage is? The amplitude is? The R.M.S voltage is?
- The sinusoidal voltage trace displayed by the CRO, the time/cm switch is on 50 µs/cm and the volt/cm switch is on 20 V/cm. The width of one complete cycle is 3.6 cm and the peak to peak height is 5.8 cm. Determine the following: The frequency is = The peak to peak voltage is = The amplitude is = The R.M.S voltage is =In an experiment, the function generator is adjusted to generate a square voltage at a certain frequency. This voltage is displayed on an oscilloscope and the reading is 10 V peak-to-peak. The rms value of this voltage is .What will be the rms value of this voltage if the frequency is reduced to the half value? 3.536 Vrms . 3.536 Vrms O 5 Vrms . 1.7677 Vrms O 3.536 Vrms. 1.7677 Vrms O 5 Vrms . 5 Vrms ...... O 10 Vrms . 10 Vrms ......A square wave having a postive maximum value of Em and a negative maximum -Em . Determine the Average Value , RMS Value , and Peak factor