Respuesta :
Answer:
Inductance of the circuit is equal to [tex]6.30\times 10^{-4}H[/tex]
Explanation:
We have given resistance R = 957 ohm
Capacitance of the circuit [tex]C=8.25\mu F=8.25\times 10^{-6}F[/tex]
Peak ac voltage is [tex]V_m=84volt[/tex]
Angular frequency at resonance [tex]\omega =88500rad/sec[/tex]
We have to find the value of inductance L
At resonance [tex]\omega L=\frac{1}{\omega C}[/tex]
[tex]\omega ^2=\frac{1}{LC}[/tex]
[tex]88500 ^2=\frac{1}{L\times 8.25\times 10^{-6}}[/tex]
[tex]L=6.30\times 10^{-4}H[/tex]
So inductance of the circuit is equal to [tex]6.30\times 10^{-4}H[/tex]
The inductance of the unknown inductor at resonance angular frequency is;
L = 1.5476 × 10^(-5) H
We are given;
Resistance; R = 957 Ω
Capacitance; C = 8.25 μF = 8.25 × 10^(-6) F
Peak voltage; V_p = 84 V
Angular frequency at resonance; ω = 88500 rad/s
In parallel circuits, resonance will occur when;
R² + ω²L² = L/C
However, it is approximated to;
ωL = 1/ωC
Making L the subject gives;
L = 1/(ω²C)
L = 1/(88500² × 8.25 × 10^(-6))
L = 1.5476 × 10^(-5) H
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