Home / UGC NET PYQ / June 2020 Solved UGC NET Electronic Science June 2020 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1.  Answer: Option (2) For Varactor Diode, m = ln(C2/C1) / ln{(1+V2/phi) / (1+V1/phi)} 2.  Answer: Option (4) 3.  Answer: Option (2) 4.  Answer: Option (2) 5.  Answer: Option (4) 6.  Answer: Option (4)  L= (5*10^-3) / P Or, P = (5*10^-3) / (5*10^7) = 10^-10 7.  Answer: Option (4) 8.  Answer: Option (3) 9.  Answer: Option (3) 10.  Answer: Option (1) 11.  Answer: Option (2) 12.  Answer: Option (2) 13.  Answer: Option (1) Slew Rate Slew rate tells us how fast the output voltage of an amplifier/op-amp can change . It is defined as: S R = max ⁡ ∣ d...

Phase Demodulation Instructions for Phase Modulation (PM) Step 1: Click on 'Generate Message' button to generate input message signal Step 2: Then click on 'Generate Carrier' button to generate carrier signal. The carrier frequency has to be more than the message frequency and You can change frequencies using sliders Step 3: Click on 'Generate Phase Modulated Signal' button to generate Phase Modulated Signal Step 4: Click on the 'Show Frequency Spectrums of PM' button to see spectrums of the PM signal Here, β represents the phase modulation index, given by β=kp*Am​, where Am​ is the amplitude of the message signal (assumed to be fixed), and kp​ is the phase sensitivity of the modulator 5 Hz Step 1: Generate Message ...

For an electricity bill, we are charged based on the amount of electrical energy consumed. 1 unit of electricity = 1 kWh (kilowatt-hour) For a constant electrical load, the energy consumed can be calculated as: Energy (kWh) = V × I × h / 1000 Where: V = Voltage in volts (V) I = Current in amperes (A) h = Operating time in hours (h) Calculation For a resistive appliance: I = V/R P = VI = V²/R At 230 V : Appliance Resistance Current Power A 100 Ω 230/100 = 2.3 A 230²/100 = 529 W B 300 Ω 230/300 = 0.767 A 230²/300 = 176.3 W So the 100 Ω appliance consumes about 3× as much power as the 300 Ω appliance. If both run for 10 hours : 100 Ω: 0.529 kW × 10 h = 5.29 kWh 300 Ω: 0.1763 kW × 10 h = 1.763 kWh Therefore, assuming the electricity tariff is the same, the 100 Ω appliance costs about 3× more to operate. ⚡ Smart Household Electricity Bill Calculator Estimate your monthly electricity bill from units consumed, calculate slab-wise charges, model household appliances, ...

1. Basic Electrical Relationship For a simple resistive electrical load, such as an ideal resistor or approximately a traditional incandescent filament: P = V × I = V / R Therefore: P = V² / R Where: P = Power in watts (W) V = Voltage in volts (V) I = Current in amperes (A) R = Resistance in ohms (Ω) 2. 5 W vs 100 W Bulb at 230 V Suppose two simple resistive bulbs operate from the same 230 V supply. 5 W Bulb R = V² / P R = 230² / 5 R = 10,580 Ω I = P / V I = 5 / 230 I ≈ 0.0217 A Therefore, the 5 W load has an effective operating resistance of approximately 10.6 kΩ and draws about 21.7 mA . 100 W Bulb R = V² / P R = 230² / 100 R = 529 Ω I = P / V ...

Charger Power Consumption & Electricity Bill Suppose the charger rating is   Input: 100–240 V AC and Output: 5 V, 1 A.  The charger's printed input/output ratings describe its operating capability.  What Does '5 V, 1 A' Actually Mean? The rating: 5 V × 1 A = 5 W means the charger can supply approximately 5 W maximum at its rated output. It does NOT mean that the charger continuously consumes 5 W. For example: Specification Meaning Input: 100–240 V The voltage range the charger can accept. Output: 5 V, 1 A The charger can provide up to approximately 5 W. Actual consumption Depends on the connected load and charger losses. The electricity meter measures energy entering the charger from the mains. ...

PM Waveform Generator Standard: Phase Modulation (PM) Message Frequency (Hz) 50 Hz Carrier Frequency (Hz) 500 Hz Kp (phase sensitivity): 50 Step 1: messagePlot $m(t) = A_m\cos(2\pi f_m t)$ Step 2: carrierPlot $c(t) = A_c\cos(2\pi f_c t)$ Step 3: superimposed $x(t) = [c(t) + m(t)]$ Step 4: modulatedPlot $s(t) = A_c\cos\left(2\pi f_c t + 2\pi k_p m(t)\right)$ Perform PM Demodulation Return to Premium Virtual DSP Lab →