Given Circuit The circuit contains: Left resistor: \(2\Omega\) Right resistor: \(2\Omega\) Vertical resistor: \(6\Omega\) Dependent current source: \(2V_2\) A, directed upward \(I_1\) and \(I_2\) enter the two-port network. Step 1: Define the Middle Node Voltage Let the voltage at the middle node be \(V_x\). Apply KCL at the middle node: $$ \frac{V_x-V_1}{2} + \frac{V_x-V_2}{2} = 2V_2 $$ Multiplying by 2: $$ 2V_x-V_1-V_2=4V_2 $$ Therefore: $$ \boxed{ V_x=\frac{V_1}{2}+\frac{5V_2}{2} } $$ Step 2: Find \(I_1\) The current entering port 1 is: $$ I_1=\frac{V_1-V_x}{2} $$ Substitute \(V_x\): $$ I_1= \frac{ V_1- \left( \frac{V_1}{2}+\frac{5V_2}{2} \ri...

  When the transformer step up voltage 230 V to 11 KV, where the current go? Because if voltage increases should not current also increase. But transformer follow energy conservation law. So,  Input Power = Output Power  Or, Vp * Ip = Vs * Is (for Ideal Transformers) where, Vp = Primary volage Vs = Secondary Voltage Ip = Primary Current Is = secondary Current So, Voltage multiplied by current on primary side equals to volage multiplied by current on secondary side. When volage is stepped up by 10 times. the current must reduce 10 times to maintain the same power. So, a transformer taking 100 amperes at 230-volt output only 10 amperes 2300-volt. That's why power plant step up volage to 400 kV for transmission because reducing current by 1000 time reduces heat loss by 1 million times in the power lines.  

GMSK Waveform Generator Standard: Gaussian Minimum Shift Keying num bits Input Bitstream Baud Rate ($R_b$) Mod. Index ($h$) Carrier (Hz) Time-Bandwidth Product (BT): Step 1: Original Message Bits Step 2: NRZ Encoding $m(t) = \sum a_n \cdot rect(t-nT)$ Step 3: Filtered Signal Plot: Gaussian Pulse Shaping $g(t) = m(t) * h_{gauss}(t)$ ...

Dual-Slope ADC Given C = 0.1 µF R = 100 kΩ V ref = 2 V V o = 10 V Integrator Equation For an integrator: V o = V ref t / RC Therefore: t = V o RC / V ref Calculate RC RC = (100 × 10 3 ) (0.1 × 10 −6 ) = 0.01 s Calculate Conversion Time t = 10(0.01) / 2 t = 0.05 s = 50 ms Therefore, t = 50 ms . Answer: B (50 ms) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Induction Motor Torque Given Frequency, f = 75 Hz Slip, s = 0.04 Rotor resistance, R 2 = 1 Ω Line voltage, V L = 415 V Phase Voltage V ph = 415 / √3 V Since stator voltage drop and rotor reactance are neglected: I 2 = V ph / (R 2 / s) Torque in Synchronous Watts P T = 3V ph 2 / (R 2 / s) P T = 3(415/√3) 2 × 1/0.04 P T = 415 2 / 0.04 ≈ 6889 W Thus, T = 6889 synchronous watts Answer: A (6889 synchronous watts) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Furth...

Transistor Power Loss Given β min = 8 R C = 11 Ω V CC = 200 V V CE(sat) = 1 V V BE(sat) = 1.5 V Overdrive factor = 5 Collector Saturation Current I C = (V CC − V CE(sat) ) / R C I C = (200 − 1) / 11 = 18.09 A Forced Beta β f = 8 / 5 = 1.6 Therefore, the base current is: I B = I C / β f I B = 18.09 / 1.6 = 11.31 A Power Loss P = V CE(sat) I C + V BE(sat) I B P = (1)(18.09) + (1.5)(11.31) P ≈ 18.09 + 16.97 = 35.06 W Therefore, P ≈ 35.07 W . Answer: A (35.07 W) Browse All Solved Papers (2012 - 2025) → ...

FDM Bandwidth Given There are 10 channels , each having a bandwidth of 50 kHz . Channel Bandwidth 10 × 50 = 500 kHz There are 10 − 1 = 9 guard bands , with each guard band having a bandwidth of 1 kHz. 9 × 1 = 9 kHz Hence, the total FDM bandwidth is: BW = 500 + 9 = 509 kHz Answer: C (509 kHz) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

VSB Transmitter Given VSB transmitted power, P VSB = 0.625 kW Modulation index, m = 0.6 Vestige = 25% = 0.25 VSB Power Equation For a VSB transmitter: P VSB = m 2 / 4 P c (1 + 0.25) Substituting the given values: 0.625 = (0.6) 2 / 4 P c (1.25) Therefore, P c = (0.625 × 4) / (0.36 × 1.25) P c = 5.56 kW Answer: B (5.56 kW) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions