Electric Circuits Advantage of active filter is do not offer gain. derive high impedance load. both of above. easy to tune. do not offer gain. derive high impedance load. both of above. easy to tune. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits For cadium plating electrolyte used is cadmium hexa metaphosphate. cadmium sulphate and sulphuric acid. sodium cyanide, cadmium and caustic soda. any of above. cadmium hexa metaphosphate. cadmium sulphate and sulphuric acid. sodium cyanide, cadmium and caustic soda. any of above. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Which of the following is not equivalent to watts? Ampere-volts. Amperes/volt. Joules per second. Ampere2 – Ohm. Ampere-volts. Amperes/volt. Joules per second. Ampere2 – Ohm. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits In a lead acid cell PbSO4 formed during both during charging as well as discharging. charging only. neither during charging nor discharging. discharging only. both during charging as well as discharging. charging only. neither during charging nor discharging. discharging only. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Consider the following circuit : Switch is closed at t = 0 Find i(0 +) 0. V / {R + (sL + 1 / Cs)}. (V*L) / (R*C) V/R. 0. V / {R + (sL + 1 / Cs)}. (V*L) / (R*C) V/R. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Constant K type HPF (high pass filter) having cut off frequency 12 KHz and nominal impedance Ro = 500 Ω Find shunt arm inductor L and series arm capacitance C for T and π section of HPF ? L = 0.0132 mH, C = 0.0132 μF. None of these. L = 3.316 mH, C = 132 μF. L = 3.316 mH, C = 0.0132 μF. L = 0.0132 mH, C = 0.0132 μF. None of these. L = 3.316 mH, C = 132 μF. L = 3.316 mH, C = 0.0132 μF. ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS