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A coil of inductance 0.50 H and resistance 100 ohm is connected to 240 V, 50Hz ac supply. What is the peak current in the coil? b) What is the time lag between the peak voltage and the peak current? |
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Answer» Solution :Here `L=0.50H, R=100ohm` `E_("rms")=240V, v=50Hz` STEP 1. We know `E_(0)=sqrt2E_("rms")=1.414xx240=339.4V` `"Impendance of LR circuit is"` `Z_(L)=sqrt(R^(2)+omega^(2)L^(2))=sqrt(R^(2)+4piv^(2)L^(2))` `sqrt((100)^(2)+4xx(3.14)^(2)xx(50)^(2)xx(0.50)^(2))="186.1 ohm"` `"PEAK CURRENT, "I_(0)=(E_(0))/(Z_(L))=(339.4)/(186.1)=1.82A` In LR circuit the phase difference between current and voltage `phi` which is given by `tan phi=(Lomega)/(R)=(Lxx2piv)/(R)` `tan phi=(0.50xx2xx3.14xx50)/(100)=1.57` `phi=tan^(-1)(1.57)=57^(0)30^(1)=57.5^(0)` `=(57.5xxpi)/(180)=0.3194 pi" radians"` `therefore "Time lag between "E_(0) and I_(0)" is given by"` `t=(phi)/(omega)=(phi)/(2piv)=(0.319pi)/(2pixx50)` `=0.003194s=3.194xx10^(-3)`s`. |
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