Capacitance C is 15.0 `muF` and the sinusoidal alternating emf device operates at amplitude `E_m` = 36.0 V and frequency `f_d`= 60.0 Hz.
(a) What are the potential difference `V_C(t)` across the capacitance and the amplitude `V_C` of `V_c(t)` ? (b) What are the current `I_C(t)` in the circuit as a function of time and the amplitude `I_C` of `I_c(t)` ?

Explanation

In a circuit with a purely capacitive load, the potential difference `v_(C)(t)` across the resistance is always equal to the potential difference E(t) across the emf device.
Calculations : Here we have `v_C(t)=E(t)` and `V_c=E_m`. Since `E_m` is given, we have
`V_C=E_m`=36.0 V.
To find `v_C(t)`,
`v_C(t)=E(t)=E_m sin omega_dt` .
Then, substituting `E_m` = 36.0 V and `omega_d=2pif_d= 120pi` , we have
`v_C=(36.0V) sin (120 pi t)`
(b) What are the current `i_C(t)` in the circuit as a function of time and the amplitude `I_C` of `i_C(t)` ? In an ac circuit with a purely capacitive load, the alternating current `i_C(t)` in the capacitance leads the alternating potential difference `v_C(t)` by `90^@` , that is , the phase constant `phi` for the current is `-90^@` , or `-pi//2` rad.
Calculation: Thus, we can write
`i_C=I_C sin (omega_d t -phi) = I_C sin (omega_d t + pi//2)`
We can find the amplitude `I_C` `(V_C=I_C X_C)` if we first find the capacitance reactance `X_C`. `(X_C=1//omega_d C)`, with `omega_d=2pif_d`, we can write
`X_C=1/(2pif_dC)=1/((2pi)(60.0Hz)(15.0xx10^(-6)F))`
=`177 Omega`.
The current amplitude is
`I_C=V_C/X_C=(36.0V)/(177Omega)=0.203` A
Substituting this and `omega_d=2pif_d=120pi` , we have
`i_C=(0.203A)sin (120pit+pi//2)`