An ac generator with emf$\mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\omega }}_{\mathbf{d}}\mathit{t}$, where${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{\text{V}}$and${\mathit{\omega }}_{\mathbf{d}}\mathbf{=}\mathbf{377}\mathbf{}\mathbf{\text{rad/s}}$, is connected to a$\mathbf{\text{4.15}}\mathbf{\text{μF}}$capacitor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{-}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{V}}$and increasing in magnitude, what is the current?

1. Amplitude of emf, ${\epsilon }_m=25.0\text{V}$
2. Angular frequency, ${\omega }_d=377\text{rad/s}$
3. Capacitance, $C=4.15\text{μF}$
4. Amplitude of emf, ${\epsilon }_m=-12.5\text{V}$

The obstruction offered in the path of alternating current by the capacitor is called capacitive reactance. Using the condition of the pure capacitive circuit, and substituting the value of capacitive reactance, we will get the maximum value of current. When the current is maximum, its first derivative is zero. Using this idea, we can find the emf of the generator. Using alternating current relation for a purely capacitive circuit, we can calculate the through the capacitor.

The voltage equation of an inductor using Ohm’s law,

${\mathit{V}}_{\mathit{C}}\mathbf{=}{\mathit{I}}_{\mathit{C}}{\mathit{X}}_{\mathit{C}}$ ...(i)

The capacitive reactance of an inductor,

${\mathit{X}}_{\mathit{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathit{\omega }}_{\mathit{d}}\mathit{C}}$ ...(ii)

The current equation for sinusoidal varying current,

${\mathit{i}}_{\mathit{L}}\left(\mathit{t}\right)\mathbf{=}{\mathit{I}}_{\mathit{L}}\mathit{s}\mathit{i}\mathit{n}\left({\mathit{\omega }}_{\mathit{d}}\mathit{t}\mathbf{-}\varphi \right)$ ...(iii)

The charge across the capacitor,

$\mathit{q}\mathbf{=}\mathit{C}\mathit{V}$ ...(iv)

For a pure capacitive circuit, the potential difference $V_C$ across the inductor is always equal to the potential difference across the emf device.

${\epsilon }_m=V_C$ ...(a)

Thus, using equation (a) and equation (iv) in equation (i), we get the maximum current value as:

$$\begin{gathered} I_C={\epsilon }_m\omega C \\ =\left(25.0\text{V}\right)\left(377\text{rad/s}\right)\left(4.15\times {10}^{-6}\text{F}\right) \\ =3.91\times {10}^{-2}\text{A} \end{gathered}$$

Hence, the value of the current is $3.91\times {10}^{-2}\text{A}$.

A varying potential difference equation across the capacitor can be given as:

$v_c=V_C\sin {\omega }_{d}t$ …(v)

$V_C$is the amplitude of the alternating voltage across the capacitor.

The charge across the capacitor is given using the above value in equation (iv) as:

$q_C=C{V}_C\sin {\omega }_{d}t$

Differentiating the charge with respect to time, we get the current across capacitor as follows:

localid="1664187002645" $$\begin{gathered} i_C=\frac{d{q}_C}{dt} \\ ={\omega }_{d}C{V}_c\cos {\omega }_{d}t \end{gathered}$$

For the current to be maximum,localid="1664187010444" $cos{\omega }_{d}t=1.$

This is possible only when

localid="1664187016586" ${\omega }_{d}t=0$

Using the above value in equation (v), the charge is given as,

localid="1664187023299" $$\begin{gathered} q_C=C{V}_C\sin \left(0\right) \\ =0 \end{gathered}$$

Hence the potential difference across the capacitor is zero, which means the emf of the generator is zero.

Emf value given in this case is exactly half the previous value and negative.

$\epsilon =-\frac{{\epsilon }_m}{2}$

And it is increasing in magnitude. Since, it already has a negative sign, an increase in its magnitude results in a more negative value.

${\epsilon }_C\left(t\right)={\epsilon }_m\sin {\omega }_{d}t$

Taking the first derivative of the above relation, we can the rate of the emf value as:

$$\begin{gathered} \frac{d}{dt}{\epsilon }_C\left(t\right)=\frac{d}{dt}\left({\epsilon }_m\sin {\omega }_{d}t\right) \\ ={\omega }_d{\epsilon }_m\cos \left({\omega }_{d}t\right) \end{gathered}$$

This result should be negative. In order to get the negative value, $cos\left({\omega }_{d}t\right)$must be negative. That implies the value as:

localid="1664187047133" $$\begin{gathered} {\omega }_{d}t=2n\pi -\frac{5\pi }{6} \\ <0 \end{gathered}$$

Substituting this value in alternating current equation (iii), we get the current value as:

localid="1664187053656" $i_C\left(t\right)=I_C\sin \left({\omega }_{d}t-\left(-\frac{\pi }{2}\right)\right)$

For the purely capacitive load, the phase constant for the current is

This gives us the further current value as:

localid="1664187060801" $$\begin{gathered} i_C\left(t\right)=I_C\sin \left(2n\pi -\frac{5\pi }{6}+\frac{\pi }{2}\right) \\ =I_C\sin \left(2n\pi -\frac{2\pi }{3}\right) \end{gathered}$$

localid="1664187068248" $\sin \left(2n\pi -\frac{2\pi }{3}\right)=\sin 2n\pi \cos \frac{2\pi }{3}-\cos 2n\pi \sin \frac{2\pi }{3}$

localid="1664187076037" $$\begin{gathered} \sin 2n\pi =0, \\ \cos 2n\pi =1, \\ \sin \frac{2\pi }{3}=-\sqrt{3}/2 \end{gathered}$$

Thus, the equation using these above equations become:

localid="1664187084002" $$\begin{gathered} i_C\left(t\right)=I_C\left(-\frac{\sqrt{3}}{2}\right) \\ =\left(3.91\times {10}^{-2}\text{A}\right)\left(-\frac{\sqrt{3}}{2}\right) \\ =-3.38\times {10}^{-2}\text{A} \end{gathered}$$

Hence, the magnitude value of the current is,localid="1664187091585" $3.38\times {10}^{-2}\text{A}$
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