In the figure below, determine the battery emf labeled ɛ1.
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the value of the required battery emf can be calculated using the KVL, the KVL can be applied to any closed path
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- What is CEMF?Determine the load resistor voltage waveform for the circuit below. Include voltage values and show computation. The rectifier is silicon. R1 +5 V 1.0 kN R1. 1.0 kN Vin oV -5 VDraw the input waveform and output waveform for the circuit given below with proper values marked in the figure. Assume D1 as germanium and D2 as silicon diodes. Input Vpp=20V, V1=5 v and V2=8 V. R DZ D1 Vin Vout V2 Maximum voltage of output waveform Minimum voltage of output waveform Windows hui
- Given the zener voltage at 9V and the zener power at 1W, what is the range of allowable values of Iz (operating values). a) 0.11A b) 22mA to 89mA c) 9A d) 0.11A to 9AA 50 ohms load resistance is connected across a half wave rectifier. The input supply voltage is 230V (rms) at 50 Hz. Determine the DC output (average) voltage (4 significant figures) Please solve it correctly. Thankyou!C4 R1 Vout For the precision rectifier circuit shown in Figure C4 what is the correct operation for the circuit if a Sine wave is 10k D1 R2 LM324 VEE applied to the input signal VIN? Vin D2 10k OUT V3 U1A Figure C4 A. When VIn is negative the circuit operates as unity voltage follower providing an in phase sine wave at VouT. When VIN is positive the circuit conducts but only to one diode drop, B. When VIn is negative the circuit operates as an inverting amplifier providing an inverted sine wave at VouT. When VIN is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, c. When VIn is negative the circuit operates as an inverting amplifier providing an in-phase sine wave at VOUT. When VIn is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, D. When VIn is negative the circuit operates as unity voltage follower providing an inverted sine wave at Vour. When VIn is positive the…
- Design a full wave rectifier based on the circuit shown in Figure 5 below: N1: N2 D3 D1 RL vs D2 DA Figure 5 The required specification for the full wave rectifier circuit includes the peak output voltage of 5 V, deliver 100 mA to the load R1, and produce an output with a ripple of not more than 5 percent. An input line voltage of 220 V (rms), 50 Hz is available. Given Vy = 0.7 V. (d) Calculate v, (max). (e) Find the required turn ratio for the transformer. (f) Determine the Peak Inverse Voltage (PIV). What is the effective load resistance? (h) Determine the required capacitance of the filter capacitor.Q3: The figure below includes a bridge full-wave rectifier circuit with a diode type is1N4007. The input voltage is Vin V2 120 sin 2n60t 5:1 D3 D1 120 V RL Vp(out) D2 DA 10 kN a) Draw the output voltage waveform for the circuit in the figure and include thevoltage values. b) What is the peak inverse voltage (PIV) across each diode? c) Determine the rms voltage, current and power delivered to RL d) Determine the average voltage, current, and power delivered to RL e) What is the ratio of Po(de) to Po(ac)? lllIn the figure below, calculate the following (use the second diode approximation): Np: N₂ 6:1 Vp 120 Vac 60 Hz (a) Vs (b) Vout(pk) (c) Vdc rü D₁ D₂ Vout R₁-2000) (d) IL (e) PIV of the diodes (f) fout
- Q2\ Determine the range of values of R₁ and I₁, that will maintain the zener diode of figure below in the on state, and determine the maximum wattage rating at the zener diode. R = 240 f www V₁60 v o- V₂=8v RL IZM = 50 mADetermine the range of V₁ that will maintain V₁ at 8V and not exceed the maximum power rating of the Zener diode. Given Rs is 88ohms and RL is 0.24 kohms. . ܫܩܦܢ V₂ o PZ man Rs V₂=8V = 400 mW R₂Power supply circuit is delivering 0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to 220V ams 5OHZ İL-DC =05A RL VL-DC =20V