Question

A capacitor is connected across an ac source that has voltage amplitude 60.0 V and frequency 80.0 Hz . (a) What is the phase angle$\mathbf{\varphi }$for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What is the capacitance C of the capacitor if the current amplitude is 5.30 A?

Short answer

The phase angle $\mathrm{\varphi }$for the source voltage relative to the current is $90°$and voltage lags the current. The capacitance C = 1.75$\times {10}^{-4}\mathbf{}\mathbf{F}$.

Step-by-step solution

Step-1: Formulas used  

$\mathbf{\varphi }$is the phase angle, the difference in the voltage and current phases.

For ac source connected with a capacitor, the capacitive reactance in the circuit is given by equation in the form

${\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{V}}{\mathbf{I}}$

Also, this capacitive reactance is related to the frequency and the capacitance C of the capacitor in the form

${\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\tau \tau fC}}$

Both equations have the same right sides, so we can get an expression for the capacitance C by

$\frac{\mathbf{V}}{\mathbf{I}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\tau \tau fC}}$, then

$\mathbf{C}\mathbf{=}\frac{\mathbf{I}}{\mathbf{2}\mathbf{\tau \tau fV}}$

Step-2: Calculation for capacitance C

Plug in the values for V, I and f to get value of C

$$\begin{gathered} \mathrm{C}=\frac{5.3\mathrm{A}}{2\mathrm{\tau \tau }\left(80\mathrm{Hz}\right)\left(60\mathrm{V}\right)} \\ =1.75\times {10}^{-4}\mathbf{F} \end{gathered}$$

The capacitor store charges and this makes the voltage lags the current and this process is opposite to the coil. So, the phase angle ϕ will be in negative sign and equals 90$°$

Therefore the phase angle $\mathrm{\varphi }$for the source voltage relative to the current is -90$°$and voltage lags the current. The capacitance C = 1.75$\times {10}^{-4}\mathbf{F}$.