Class 12 Physics Alternating Current | Power associated with AC circuit |

**Power associated with AC circuit**

Consider source voltage V = V_{m} sinωt ,

- I = I
_{m}V_{m}sin( ωt + f)

where f = phase angle between current and voltage.

- I
_{m}= (V_{m}/Z) and f = tan^{-1}(X_{C}–X_{L})/(R) - Instantaneous power p = VI = V
_{m}I_{m}sin ωt sin(ωt + f) - p = (V
_{m}I_{m}/2) [cos(-f) - cos(2 ωt + f)] [ By using 2sinA sinB = cos(A-B) cos(A+B)] **p =****(****V**_{m}I_{m}/2) [cos**f****- cos(2****ωt +****f****)]**- Average Power P = (1/T)
_{0}^{T}∫ p dt - = (1/T)
_{0}^{T}∫ (V_{m}I_{m}/2) [cosf - cos(2 ωt + f)] dt

After Simplifying ,

**P = (****V**_{m}I_{m}/2) cos**f**- Where cosf =
__power factor__.

__Power in different AC circuits__

- Resistive :-
- f = 0 because voltage and current are in phase.
- Therefore P = (V
_{m}I_{m}/2). There will be maximum power dissipation. - Inductive:-
- f = (∏/2) as current lags behind the voltage by (∏/2) .
- Therefore P = 0.
- Capacitive:-
- f = (∏/2) as voltage lags behind the current by (∏/2) .
- Therefore P =0.
- LCR:-
- f = tan
^{-1}(X_{C}–X_{L})/(R) - Power dissipates only in resistor.
- At resonance in LCR :-
- X
_{C}= X_{L} - Therefore f = 0.
- P = (V
_{m}I_{m}/2). There will be maximum power dissipation.

** Problem:- **A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.

(a) Obtain the current amplitude and rms values.

(b) Obtain the rms values of potential drops across each element.

(c) What is the average power transferred to the inductor?

(d) What is the average power transferred to the capacitor?

(e) What is the total average power absorbed by the circuit?

[‘Average’ implies ‘averaged over one cycle’.]

__Answer:-__

Inductance, L = 80 mH = 80 × 10^{−3} H

Capacitance, C = 60 μF = 60 × 10^{−6} F

Supply voltage, V = 230 V

Frequency, ν = 50 Hz

Angular frequency, ω = 2πν= 100 π rad/s

Peak voltage, V_{0} = V √2

(a) Maximum current is given as:

Hence, RMS value of current,

I_{0} = (V_{0})/( ωL – (1/ ωC))

=(230 √2)/((100 π x 80 x10^{-3}) –(1/(100 π x60 x 10^{-6}))

=(230 √2) / ( 8π x (1000/6 π)) =-11.63A

The negative sign appears because (ωL) < (1/ ωC).

Amplitude of maximum current, II_{0}I =11.63A

I= (I_{0})/( √2)

=(-11.63)/( √2)

=-8.22A

(b) Potential difference across the inductor,

V_{L}= I × ωL

= 8.22 × 100 π × 80 × 10^{−3}

= 206.61 V

Potential difference across the capacitor,

V_{C} = (I x (1/ ωC))

= (8.22) x (1/ 100 π x 60 x 10^{-6})

=436.3V

The negative sign appears because

Amplitude of maximum current,

(c) Average power consumed by the inductor is zero as actual voltage leads the current by π/2.

(d) Average power consumed by the capacitor is zero as voltage lags current by π/2.

(e) The total power absorbed (averaged over one cycle) is zero.

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