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A coil of 0.01 henry inductance and 1Omega resistance is connected to 200 V, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current. |
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Answer» SOLUTION :Here , L=0.01 H, R=1 `Omega` ,V=200 V Frequency v=50 Hz Impendance of the circuit , `Z=sqrt(R^2+X_L^2)=sqrt(R^2+(2pivL)^2)` `therefore Z=sqrt((1)^2+(2xx3.14xx50xx0.01)^2)` `=sqrt(1+9.8596)=sqrt(10.8596)` `therefore Z=3.295 Omega` `therefore Z approx 3.3 Omega` `TAN phi =(omegaL)/R=(2pivL)/R` `therefore tan phi=(2xx3.14xx50xx0.01)/1` `therefore tan phi =3.14` `therefore phi = tan^(-1) (3.14)` `therefore phi =72^@` `=(72xxpi)/180` rad Phase DIFFERENCE , `phi=omegaDeltat` `therefore Deltat=phi/omega=(72pi)/(180xx2pivL)=72/(360xx50xx0.01)` `therefore Deltat=1/250` s |
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