Electric Circuits A 230V, 50Hz voltage is applied to a coil having L = 5H and R = 2 Ohm in series with capacitance C. When the voltage across the coil is 400V, find the current in the circuit? 0.4583 A. 0.1578 A. 0.1465 A. 0.2547 A. 0.4583 A. 0.1578 A. 0.1465 A. 0.2547 A. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Which property is used to oppose the flow of current ? Capacitance Inductance Poteintial difference Resistance Capacitance Inductance Poteintial difference Resistance ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Inductor does not allow the sudden change of power current None of these voltage power current None of these voltage ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Condition of symmetry in Z-parameter representation is Z11 = Z22. Z12 = Z21. Z12 = Z22. Z11 = Z12. Z11 = Z22. Z12 = Z21. Z12 = Z22. Z11 = Z12. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Disadvantages of constant k type filter none of these. both A and B. characteristic impedance unchanged in pass band. attenuation does not increase rapidly beyond cut off frequency. none of these. both A and B. characteristic impedance unchanged in pass band. attenuation does not increase rapidly beyond cut off frequency. ANSWER DOWNLOAD EXAMIANS APP
Electric Circuits Open circuit voltage of a fully charged lead acid cell is 2.7 V. 1.9 V. 2.5 V. 2 V. 2.7 V. 1.9 V. 2.5 V. 2 V. ANSWER DOWNLOAD EXAMIANS APP
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