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प्रश्न
An alternating current I = 14 sin (100 πt) A passes through a series combination of a resistor of 30 Ω and an inductor of `(2/(5pi))` H. Taking `sqrt2` = 1.4 calculate the power factor of the circuit.
उत्तर
Given: I = 14 sin (100 πt) A, R = 30 Ω, L = `2/(5pi)`H
From here
ω = 100π
`X_L = omegaL` (inductive reactance)
= `100pi xx 2/(5pi) = 40Omega`
Impedance, Z = `sqrt(R^2 + X_L^2) = sqrt((30)^2 + (40)^2)`
Z = `sqrt(900 + 1600) = 50Omega`
Power factor = `cosphi`
phasor diagram for finding `phi`,
`cosphi = R/Z`
= `30/50`
`cosphi = 3/5`
Power factor = 0.6
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