The operating pressure is 2.1 MPa and the pump efficiency is 60%. The nozzle spacing and forward speed are 400 mm and 3.4 km h -1, respectively. The application rate of an 18-nozzle hydraulic sprayer is 1120 L ha-1. The power in ?(?) in watts isĪ good transconductance amplifier should have -gate Electronics and Communication 2017
The voltage across and the current through a load are expressed as follows v(t) = −170 sin (377t − π/6 ) V i(t) = 8 cos (377t + π/6 ) A The average power in watts (round off to one decimal place) consumed by the load is _.Ĭ onsider a white Gaussian noise process ?(?) with two-sided power spectral density ??(?) = 0.5 W/Hz as input to a filter with impulse response 0.5?−?2/2 (where ? is in seconds) resulting in output ?(?). output of a rectifier (Physics CBSE 2018) (b) a zener diode works to obtain a constant d.c. (Physics CBSE 2018)Įxplain briefly how (a) a barrier potential is formed in a p-n junction diode. Does it violate the law of conservation of energy ? Justify your answer. If switch-on duration is 1 ms, the load power is (A) 6 W (B) 12 W (C) 24 W (D) 48 WĪ step-up transformer converts a low input voltage into a high output voltage. The switching frequency of the converter is 250 Hz. It has 48 V input voltage, and it feeds a resistive load of 24 Ω. The value of the boost inductor (in μH) is _.Ī DC-DC buck converter operates in continuous conduction mode. The converter operates at the boundary between continuous and discontinuous conduction modes. Assume ideal components and a very large output filter capacitor. In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The additional load at unity power factor in kW (round off to two decimal places) that may be added before this transformer exceeds its rated kVA is _. GATE EC 2015Ī single-phase transformer of rating 25 kVA, supplies a 12 kW load at power factor of 0.6 lagging. The maximum power dissipation in Q1 (in Watts) is _.
#Bridge rectifier calculator eight diode full#
The percentage regulation of the transformer at full load with 0.8 lagging power factor is
#Bridge rectifier calculator eight diode series#
If R L is 1k Ω and forward resistanc e of diode is 50 Ω, find the followings:Ī single-phase 100 kVA, 1000 V / 100 V, 50 Hz transformer has a voltage drop of 5% across its series impedance at full load. The i/p to the Full wave rectifier is v(t) = 200 sin50t. Calculate the average voltage, DC output power, ac input power, rectification efficiency and p The load and diode forward resistances are 100 Ω and 10 Ω respectively. The term "diode resistance" is equal to "forward diode resistance" in this passage.In a full wave rectifier, the input is from 30-0-30 V transformer. Please notify the admin if there are any bugs or requests. The output voltage and efficiency is rounded to the hudredths place. Thus, the efficiency cannot exceed that value the rectifying efficiency decreases as the diode resistance increases. The rectifying efficiency of a half wave rectifier when the forward diode resistance is zero is approximately 40.53%. Since the half wave rectifier only rectifies half of a sin wave, the average ouput voltage is half of that of a full wave rectifier in the same conditions. It is also affected by the load resistance when the diode resistance is not zero. This value decreases as the diode resistance increases. The average output voltage of a half wave rectifier when the diode resistance is zero is approximately 0.318*AC Input Voltage(max)) or 0.45*AC Input Voltage(RMS). This tool calculates the average output voltage and rectifying efficiency of a half wave rectifier while taking into account the forward diode resistance. Half Wave Rectifier - Average Output Voltage and Rectifying Efficiency Calculator