For the circuit shown in figure, the impedance of the circuit will be

Question

For the circuit shown in figure, the impedance of the circuit will be (A)  50Ω  (B)  60Ω  (C)  90Ω  (D)  120Ω

Tardigrade Admin · Accepted Answer

Current through inductance, iL=(90/30)=3A Current through capacitor, iC=(90/20)=4.5A Net current through circuit i=iC-iL=4.5-3=1.5A ∴ Impedance of the circuit, Z=(V/i)=(90/1.5)=60Ω

Q. For the circuit shown in figure, the impedance of the circuit will be

$50Ω$

$60Ω$

$90Ω$

$120Ω$

Current through inductance, $i_{L} = \frac{90}{30} = 3 A$
Current through capacitor, $i_{C} = \frac{90}{20} = 4.5 A$
Net current through circuit $i = i_{C} - i_{L} = 4.5 - 3 = 1.5 A$
$∴$ Impedance of the circuit,
$Z = \frac{V}{i} = \frac{90}{1.5} = 60Ω$