Calculate Total Circuit Admittance at 550 Hz ad 1000 Hz Solution a>impedence of

Question

Calculate Total Circuit Admittance at 550 Hz ad 1000 Hz

Explanation / Answer

a>impedence of resistor=470 impedence of inductor=jwL=j2*3.14*550 *47*10^-3=j 162.34 impedence of capacitor=-j/wC=-j /2*3.14*550*10^-6=-j289.52 1/Zeq=1/Z1+1/Z2+1/Z3=1/470+1/j162.34 +1/-j289.52 =>Zeq=179.56 + j228.36 admittance=1/impedence=2.12*10^-3-j2.70*10^-3 b>impedence of resistor=470 impedence of inductor=jwL=j2*3.14*1000 *47*10^-3=j 295.16 impedence of capacitor=-j/wC=-j /2*3.14*1000*10^-6=-j159.23 1/Zeq=1/Z1+1/Z2+1/Z3=1/470+1/j295.16 +1/-j159.23 =>Zeq=165.03+j-224.34 admittance=1/impedence=2.12*10^-3+j2.89*10^-3

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