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Q.Derive an expression for the power factor of an alternating circuit having inductance and resistance in series.

Odisha ChseOdisha CHSE +2 Science Board Exam 2025Subjective· 3mImportance★★★★★
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Impedance triangle for a series L-R circuit: R (base), X_L = ωL (vertical), Z (hypotenuse), phase angle φ, giving cosφ = R/Z.
Impedance triangle for a series L-R circuit: R (base), X_L = ωL (vertical), Z (hypotenuse), phase angle φ, giving cosφ = R/Z.

cos(phi) = R/Z = R/sqrt(R^2 + (wL)^2) for a series L-R circuit.

In a series circuit with resistance R and inductance L driven by an AC source, the voltage across R (V_R = IR) is in phase with the current, while the voltage across L (V_L = I X_L, XL=ωLX_L = \omega L) leads the current by 90∘90^\circ. Using a phasor diagram, these add at right angles, so the total voltage is

V=VR2+VL2=IR2+XL2.V = \sqrt{V_R^2 + V_L^2} = I\sqrt{R^2 + X_L^2}.

The impedance is therefore Z=R2+XL2=R2+(ωL)2Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (\omega L)^2}.

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