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Question
The variation of inductive reactance (XL) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.
- Calculate the self-inductance of the inductor.
- When this inductor is used in series with a capacitor of unknown value and a resistor of 10 Ω at 300 s–1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
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Solution 1
i. We know that XL = ωL and ω = 2π f
Where,
f = frequency in Hz
So, L = `X_L/(2 pi f)`
= `(20)/(2 pi(100))`
= `(40)/(2pi (200))`
= `(60)/(2pi(300))`
= 31.84 × 10−3
≈ 32 mH
ii. We know that power dissipation is maximum when,
XL = XC
⇒ ωL = `1/(omegaC)`
⇒ `C = 1/(omega^2L)`
⇒ C = `1/(4 pi^2 f^2 L)`
= `1/(4 xx 3.14 xx 3.14 xx 300 xx 300 xx 32 xx 10^-3)`
= 8.8 μF
Solution 2
i. Given: From graph f = 100 Hz
XL = 20 Ω
Inductive reactance (XL) = 2π f L
So,
L = `X_L/(2 pi f)`
= `(20)/(2 pi xx 100)`
= 0.032 H
= 32 mH
ii. Given: f = 300 s−1
L = 0.032H
We know that power dissipation is maximum when,
2π f L = `1/(2 pi f C)`
2π × 300 × 0.032 = `1/(2 pi xx 300 xx C)`
∴ C = 8.8 × 10−6 F
= 8.8 μF
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