induebo VA 3 SolutionFOR THE GIVEN CIRCUIT total impedanceZ3

induebo VA 3

Solution

FOR THE GIVEN CIRCUIT

total impedance=Z=(3+4i)(-5i)/(3+4i-5i)

Z=20-15i/(3-i)

Z=7.9 ang-18.4

current I=V/Z

I =25/7.9 ang(-18.4)

I=3.16 NG(18.4)

Current through inductor= I*(-5I)/(3-I)

=3.16 ang(18.4)*(-5i)/(3-i)

IL =4.99 ang(-53.1)

voltage across inductor=IL* ZL

   VL=4.99 ang(-53.1) *4i

   VL= 19.96 ang (36.9) V

 induebo VA 3 SolutionFOR THE GIVEN CIRCUIT total impedance=Z=(3+4i)(-5i)/(3+4i-5i) Z=20-15i/(3-i) Z=7.9 ang-18.4 current I=V/Z I =25/7.9 ang(-18.4) I=3.16 NG(1

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