Question

What is the impedance (in Ω) of a series combination of a 68 Ω resistor, a 1.5 µF capacitor, and a 3 µF capacitor at a frequency of 3.0 kHz? _____Ω

Answer 1

Sol::

Given
Resistance(R) = 68 Ω
Capacitance (C1) = 1.5 µF
C2 = 3 µF
frequency(f) = 3000 Hz

As capacitors are in series, their equivalent be

Ceq= (C1*C2)/(C1+C2)
= 1.5*3*10^-12/(1.5+3*10^-12)
= 1*10^-6 F

Now we have impedance (Z) = √(R²+(xl-xc)²)

Where
XL= 0

And

Xc = 1/2πfCeq
= 1/(2π*3000*1*10^-6)
= 53.06 Ω

So

Z= √[68²+53.06²]
= 86.28 Ω

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