In an a.c circuit, value of voltage and current change every instant. Therefore, power of an a.c. circuit at any instant is the product of instantaneous voltages `(E)` and instantaneous current `(I)`. The average power supplied to a pure resistance `R` over a complete cycle of a.c is `P = E_(upsilon). I_(upsilon)`. When circuit is inductive, average power//cycle `= E_(upsilon) I_(upsilon) cos phi`, where `phi` is the phase angle between alternating voltage an altenating current in the circuit.
In an a.c. circuit, `800 mH` inductor and a `60 mu F` capacitor are connected in series with `15 ohm` resistance. The a.c. supply to the circuit is `230 V, 50 Hz`
This average power transferred per cycle to resistance is

Options

A. `2.05 W`
B. `20.05 W`
C. Zero
D. `230 W`

Answer

B

Explanation

Here, `L = 500 mH = 800 xx 10^(-3) H = 0.8 H`
`C = 60 mu F = 60 xx 10^(-6) F`
` R = 15 ohm, E_(upsilon) = 230 V` and `v = 50 Hz`.
`X_(L) = omega L = 2 pi v L`
`2 xx 22//7 xx 50 xx 0.8 = 251.43 Omega`
`X_(C ) = (1)/(omega C) = (1)/( 2 pi v C) = (1)/(2 xx 22//7 xx 50 xx 60 xx 10^(-6))`
`= 53.03 ohm`
`Z =sqrt(R^(2) + (X_(L) - X_(C ))^(2)) = sqrt(15^(2) + (251.43 - 52.03)^(2))`
`= 98.98 ohm`
`I_(upsilon) = (E_(upsilon))/(Z) = (230)/(198.97) = 1.156 A`
`:.` Average power transferred/cycle to resistance
`P_(R ) = I_(upsilon)^(2) xx R = (1.156)^(2) xx 15 = 20.05` watt