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Question:-

An A.C circuit having an inductor and a resistor in series draws a power of 560 W from an A.C source marked 210 V - 60 Hz. The power factor of the circuit is 0.8. Calculate the impedance of the circuit and the inductance of the inductor.

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Answer 1

Answer:-

$\red{\bigstar}$ The impedance of the circuit is $\large\leadsto\boxed{\rm\purple{63 \: \Omega}}$

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$\red{\bigstar}$ The inductance of the inductor is $\large\leadsto\boxed{\rm\purple{0.1 \: H}}$

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• Solution:-

➊

We know,

The average power over a complete cycle is given by

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{P = E_{rms} \times i_{rms} \times cos \varnothing}}}$

here,

• cos ∅ is the power factor.

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➪ $\sf i_{rms} = \dfrac{P}{E_{rms} \times cos \varnothing}$

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➪ $\sf i_{rms} = \dfrac{560 \: W}{210 \: V \times 0.8}$

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➪ $\sf \dfrac{10}{3} A$

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✯ The impedance of the circuit is

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{Z = \dfrac{E_{rms}}{i_{rms}}}}}$

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➪ $\sf \dfrac{210 \: V}{(10/3)A}$

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★ $\large{\underline{\underline{\bf\red{63 \: \Omega}}}}$

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➋

✯ The power is consumed in R only. Therefore,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{P = (i_{rms})^2 \: R}}}$

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➪ $\sf R = \dfrac{P}{(i_{rms})^2}$

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➪ $\sf R = \dfrac{560 \: W}{(\frac{10}{3} A)^2}$

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★ $\large\bf{50.4 \: \Omega}$

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✯ We know, the impedance of an L-R Circuit is

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$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{Z = \sqrt{R^2 + (\omega L)^2}}}}$

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➪ $\sf (\omega L)^2 = Z^2 - R^2$

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➪ $\sf (\omega L)^2 = (63 \: \Omega)^2 - (50.4 \: \Omega)^2$

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➪ $\sf (\omega L)^2 = (63 + 50.4) \Omega \times (63 - 50.4) \Omega$

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➪ $\sf (\omega L)^2 = 113.4 \Omega \times 12.6 \Omega$

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➪ $\sf (\omega L)^2 = 1428.84 (\Omega)^2$

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➪ $\sf (\omega L) = \sqrt{1428.84}$

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➪ $\sf (\omega L) = 37.8 \: \Omega$

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➪ $\sf L = \dfrac{37.8 \: \Omega}{2 \pi f}$

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➪ $\sf L = \dfrac{37.8 \Omega}{2 \times 3.14 \times 60 \: s^{-1}}$

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★ $\large{\underline{\underline{\bf\red{0.1 \: H}}}}$

Answer 2

Given :-

An A.C circuit having an inductor and a resistor in series draws a power of 560 W from an A.C source marked 210 V - 60 Hz. The power factor of the circuit is 0.8

To Find :-

Impedance

Inductance

Solution :-

We know that

$\sf \bar{P} =V_{rms}\times I_{rms}\times cos(\theta)$

$\sf 560=210\times I_{rms}\times 0.8$

$\sf 560=168\times I_{rms}$

$\sf\dfrac{560}{168}=I_{rms}$

$\sf 3.33=I_{rms}$

Now

$\sf Impedane=\dfrac{V_{rms}}{I_{rms}}$

$\sf Impedane=\dfrac{210}{3.33}$

$\sf Impedane=\dfrac{2100}{333}$

$\sf Impedane\approx 63\;\Omega$

$\sf P=(I_{rms})^2\times R$

$\sf 560=3.33(R)$

$\sf\dfrac{560}{3.33}=R$

$\sf\dfrac{5600}{333}=R$

$\sf 50.4\;\Omega$

Now

$\sf Z^2=R^2+\omega L^2$

$\sf 63^2=50.4^2-\omega L^2$

$\sf 3969=2540.16+\omega L^2$

$\sf 3969-2540.16=\omega L^2$

$\sf 1428.84=\omega L^2$

$\sf \sqrt{1428.84} =\omega L^2$

$\sf 37.8=\omega L$

Now

$\sf Inductance = \dfrac{\omega L}{2\pi f}$

$\sf Inductance = \dfrac{37.8}{2\times\dfrac{22}{7}\times60}$

$\sf Inductance = \dfrac{37.8}{\dfrac{44}{7}\times60}$

$\sf Inductance = \dfrac{37.8}{377}$

$\sf Inductance = 0.1\;H$