In the diagram shown in figure, `V` function is given. Find other four functions of time `I, V_C, V_R` and `V_L`. Also, find power consumed in the circuit, `V` is given in volts and `omega` in rad/s.

Explanation

Given `omega=100rad//s`
`X_L=omegaL=50Omega`
`X_C=1/(omegaC)=1/(100xx10^-3)=10Omega`
`Z=sqrt(R^2+(X_L-X_C)^2)`
`=sqrt((30)^2(50-10)^2)`
`=50Omega`
current function
Maximum value of current
`I_0=V_0/Z=200/50=4A`
`X_LgtX_C`, therefore voltage leads the current by a phase difference `phi` where,
`cosphi=R/Z=30/50=3/5`
`or phi=53^@`
`:. I=4sin(100t+30^@-53^@)`
`or I=4sin(100t-23^@)`
`V_R,V_C` and `V_L` functions
maximum value of `V_R=I_0R=4xx30=120` volt, `V_R` and `I` are in same phase.
Therefore,
`V_R=120sin(100t-23^@)`
Maximumk value of `V_C=I_0X_C+4xx10=40`volt
now, `V_C` function lags the current function by` 90^@`
Therefore
`V_C=40sin(100t-23^@-90^@)`
or `V_C=40sin(100t-113^@)`
Maximum value of `V_L=I_0X_L=4xx50=200` volt, `V_L` function leads the current function by `90^@`.
Therefore
Maximum value of `V_L=200sin(100t-23^@+90^@)`
or `V_L=200sin(100t+67^@)`
Power is consumed in an `AC` circuit any across reistance and this power is given by `P=V_("rms")I_("rms")cos phi`
Let us use the first formula
`P=(200/sqrt(2))(4/sqrt(2))(3)/(5)`
`=240` watt.