Question

(a) In a series L-C-R circuit with an AC source, R =` 300 Omega`, C = `20 muF, L= 1.0 H, V_(0) = 50sqrt2V and f = 50/pi Hz`. Find (i) the rms current in the circuit and (ii) the rms voltage across each element.
(b) Consider the situatiuon of the previous part. find the average electric field energy stored in the capacitor and the average magnetic field energy stored iun the coil .

Answer 1

Consider the series L-C-R circuit as shown below

It is given that `V_(O) = 50sqrt(2)V` ltbgt and `omega=2pif = 2pi xx 50/pi = 10 rad s^(-1)`
`therefore X_(L) = omegaL= 100 xx 1 = 100 Omega`
`therefore X_(C) = 1/(omegaC) = (1)/(100 xx 20 xx 10^(-6)) = 500 Omega`
`rArr Z= sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`=sqrt((300)^(2)+(500-100)^(2))` `=500Omega`
`therefore` Peak value of current `I_(0) = V_(0)/Z = (50sqrt(2))/500 = 0.1 sqrt(2)A`
i) `therefore I_(rms) = I_(0)/sqrt(2) = (0.1sqrt(2))/sqrt(2) = 0.1 A`
ii) rms voltage across each element is
`V_(R) = I_(rms)R= 0.1 xx 300 = 30V`
`V_(L) = I_(rms)X_(L) = 0.1 xx 100 = 10V`
and `V_(C) = I_(rms)X_(C) = 0.1 xx 500=5V`
b) The average electric field energy stored in capacitor is given by
`U_(C) = 1/2CV_(C)^(2) = 1/2 xx 20 x 10^(-6) xx (50)^(2) = 25 mJ`
also the average magnetic field energy stored in the coil is given by
`U_(L) = 1/2LI_(rms)^(2) = 1/2 xx 1 xx (0.1)^(2)=5mJ`