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A 50 Hz AC signal is applied in a circuit of inductance of `(1//pi)`H and resistance `2100Omega.` The impedance offered by the circuit isA. `1500 Omega`B. `1700 Omega`C. `2102 Omega`D. `2500 Omega` |
Answer» Correct Answer - C Impedance, `Z= sqrt(R^(2)+X_(L)^(2)), X_(L)=omegaL=2pifL` Given, `R=2100Omega`, f=50Hz, `L=1/piH` `rArr Z= sqrt((2100)^(2) + (2 xx 50)^(2)) = sqrt((2100)^(2) + (100)^(2))` `rArr Z=2102 Omega` |
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