Question 9.86: The circuit shown in Fig. 9.86 is used in a television recei...

The circuit shown in Fig. 9.86 is used in a television receiver. What is the total impedance of this circuit?

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Y=1240+1j95+1j84Y=4.1667×103j0.01053+j0.0119Z=1Y=10004.1667+jl.37=10004.386118.2Z=22818.2Ω\begin{array}{l}{Y}=\frac{1}{240}+\frac{1}{j 95}+\frac{1}{-j 84} \\\\{Y}=4.1667 \times 10^{-3}-\mathrm{j} 0.01053+\mathrm{j} 0.0119 \\\\{Z}=\frac{1}{{Y}}=\frac{1000}{4.1667+\mathrm{jl} .37}=\frac{1000}{4.3861 \angle 18.2^{\circ}} \\\\{Z}={228} \angle {- 1 8 . 2}^{\circ} {\Omega}\end{array}

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