First slide

AC Circuits

Question

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

Difficult

A

$65 Ω, 0.2 H$
B

$64 Ω, 1.0 H$
C

$63 Ω, 0.1 H$
D

$50 Ω, 1.5 H$

Solution

$T h e a v e r a g e p o w e r o v e r a c o m p l e t e c y c l e i s P = E_{rms} \times I_{rms} \times \cos ϕ ∴ I_{rms} = \frac{P}{E_{rms} \times \cos ϕ} = \frac{560}{210 \times 0.8} = \frac{10}{3} A The impedance of the circuit is Z = \frac{E_{rms}}{I_{rms}} = \frac{210 V}{(10 / 3) A} = 63 Ω Power factor = cosϕ = \frac{R}{Z} \Rightarrow R = 0.8 \times 63 = 50.4 Ω Now, the impedance of an L - R circuit is Z = \sqrt{R^{2} + (ωL)^{2}} ∴ (ωL)^{2} = Z^{2} - R^{2} = (63)^{2} - (50.4)^{2} = 1428.84 Ω^{2} \Rightarrow ωL = \sqrt{1428.84} = 37.8 Ω ∴ L = \frac{37.8}{2 πf} = \frac{37.8}{2 \times 3.14 \times 60} = 0.1 H$

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