SSE JE Electrical Engineering 28 Oct 2020 Shift 1 Papers

Question : 152 of 200

Marks: +1, -0

An R-L-C series resonant circuit has the following parameters: Resonance frequency = 5000 / 2π Hz; impedance at resonance = 56 Ω and Q-factor = 25. Calculate the capacitance of the capacitor.

143 μF
0.143 μF
1.43 μF
14.3 μF

Validate

Solution:

Resonance frequency

('\f)=5000 / {2 \π}\Hz

Resonant frequency

(\Ω)=2 \π\f =5000 \rad \/ \sec

Impedance at resonance

( \Z )= \R =56 \Ω

Quality factor

=25

\Q ={\ω\L }/{ \R }

⇒25={5000 ×\L }/{56}

⇒\L =0.28 \H

\Q ={1}/{ \R } √{{ \L }/{ \C }}

⇒ 25={1}/{56} √{{0.28}/{ \C }}

⇒\C =0.143 \µ\F

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