In a series, L.C.R circuit an alternating emf (v) and current (i) are given by the equation v = v₀ sin (ω t), $i = i_{0} \sin (ω t + \frac{π}{3})$
The average power dissipated in the circuit over a cycle of AC is

Question

Question

Answer 1

CONCEPT:

The ac circuit containing the capacitor, resistor and inductor is called LCR circuit.
The root mean square current /voltage (rms) of an AC circuit is the effective current/voltage of that circuit.
The maximum value of the potential/current in an Ac circuit is called the peak value of voltage/current.

The average power dissipated in an AC circuit is given by:

P_avg = V_rmsI_rmsCosθ

Where V_rmsis rms potential, I_rmsis rms current and θ is phase difference between potential and current

$r m s c u r r e n t (I_{r m s}) = \frac{I}{\sqrt{2}}$

$r m s v o l t a g e (V_{r m s}) = \frac{V}{\sqrt{2}}$

Where I is maximum/peak current in the circuit and V is peak voltage

EXPLANATION:

Given that:

Potential (v) = v₀ sin (ω t)

$C u r r e n t (i) = i_{0} \sin (ω t + \frac{π}{3})$

Phase difference (θ) = π/3

Peak current (I) = i₀

Peak voltage (V) = v₀

$r m s c u r r e n t (I_{r m s}) = \frac{I}{\sqrt{2}} = \frac{i_{0}}{\sqrt{2}}$

$r m s v o l t a g e (V_{r m s}) = \frac{V}{\sqrt{2}} = \frac{v_{0}}{\sqrt{2}}$

Use formula

P_avg = V_rmsI_rmsCosθ

A v e r a g e p o w e r (P_{a v g}) = \frac{v_{0}}{\sqrt{2}} \times \frac{i_{0}}{\sqrt{2}} \times C o s \frac{π}{3} = \frac{v_{0} i_{0}}{2} \times \frac{1}{2} = \frac{v_{0} i_{0}}{4}