I=1.0A as givenVR=160V=IRR=I×R1×160=160VL=120V=IRR=I×R1×120=120Vnet=V2R+V2L=1602+1202=25600+14400=40000=200effective value of applied voltage=200Z=R2+L2=1602+1202=25600+14400=40000=200impedance=200I =1.0A\space as\space given\\ V_R=160V=IR R=I\times R 1\times 160=160\\ V_L=120V=IR R=I\times R 1\times 120=120\\ V_{net}= \sqrt {V ^{2}{_R}+ {V ^{2}{_L} }}\\ \qquad=\sqrt {160 ^{2}+ {120 ^{2}}}\\ \qquad=\sqrt {25600+ 14400}\\ \qquad =\sqrt {40000}\\ \qquad=200\\ \qquad effective \space value\space of \space applied\ voltage=200\\ Z= \sqrt {R ^{2}+ {L^{2} }}\\ \qquad = \sqrt {160 ^{2}+ {120^{2} }}\\ \qquad=\sqrt {25600+14400}\\ \qquad=\sqrt {40000}\\ \qquad=200\\ impedance=200I=1.0A as givenVR=160V=IRR=I×R1×160=160VL=120V=IRR=I×R1×120=120Vnet=V2R+V2L=1602+1202=25600+14400=40000=200effective value of applied voltage=200Z=R2+L2=1602+1202=25600+14400=40000=200impedance=200
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