Learn capacitor in an AC circuit and calculate the current as a function of time equation. Practice to find phase difference between voltage & current in purely capacitive circuit.

- Consider an AC circuit in which only a capacitor is connected with AC voltage.
- When only capacitor is connected across battery, then such circuit is known as purely capacitive circuit.
- The capacitance of capacitor is C.
- The voltage across capacitor is \(V=V_0\,sin\,\omega t\).
- Charge on capacitor

Applying Kirchhoff's law-

\(V-\dfrac{q}{C}=0\)

\(V_0\,sin\,\omega t=\dfrac{q}{C}\)

\(q=C\,V_0\,sin\,\omega t\)

Here, q is the instantaneous charge on the capacitor

- Current across capacitor

Differentiating the equation of charge on capacitor

\(\dfrac{dq}{dt}=i_c=\omega\, CV_0\,cos\,\omega t\)

\(\because\,cos\,\theta=sin\left(\dfrac{\pi}{2}+\theta\right)\)

\(\therefore \,i_C=\omega\, C V_0\,sin\left(\dfrac{\pi}{2}+\omega t\right)\)

A \(0.9\,sin\,120\,\pi t\)

B \(0.45\,cos\,120\,\pi t\)

C \(10\,sin\,\,120\,\pi t\)

D \(15\,cos\,\,120\,\pi t\)

- Consider a purely capacitive circuit, as shown in figure.

- The voltage source is given by

\(V=V_0\,sin\,\omega t\)

- The current is given by

\(I_C=\omega \,CV_0\,sin(\dfrac{\pi}{2}+\omega t)\)

1) Convert the equation of current in terms of cosine.

\(I_C=\omega\, CV_0\,cos\,\omega t\)** **

2) At time, \( t=0\)

\(I_C=\omega\, CV_0\,cos\,\omega (0)\)

\(I_C=\omega \,CV_0\)

3) At time, \(t=\dfrac{T}{4}\)

\(\because\,\dfrac{T}{4}=\dfrac{2\pi}{4\omega}=\dfrac{\pi}{2\omega}\)

\(\therefore\,I_C=\omega CV_0\,cos\,\omega\left(\dfrac{\pi}{2\omega}\right)\)

\(I_C=0\) ** **

4) At time, \(t=\dfrac{T}{2}\)

\(\because\,\dfrac{T}{2}=\dfrac{1}{2}\left(\dfrac{2\pi}{\omega}\right)=\dfrac{\pi}{\omega}\)

\(\therefore\,I_C=\omega\, CV_0\,cos\,\omega\left(\dfrac{\pi}{\omega}\right)\)

\(I_C=-\omega\, CV_0\)

5) At time, \(t=\dfrac{3T}{4}\)

\(\because\,\dfrac{3T}{4}=\dfrac{3}{4}\left(\dfrac{2\pi}{\omega}\right)=\dfrac{3\pi}{2\omega}\)

\(\therefore\,I_C=\omega \,CV_0\,cos\,\omega\left(\dfrac{3\pi}{2\omega}\right)\)

\(I_C=0\)** **

6) At time, \(t=T\)

\(\because \,T=\dfrac{2\pi}{\omega}\)

\(\therefore\,I_C=\omega\, CV_0\,cos\,\omega\left(\dfrac{2\pi}{\omega}\right)\)

\(I_C=\omega \,CV_0\)

- Consider a purely capacitive circuit, as shown in figure.

Equation of voltage : \(V=V_0\,sin\,\omega t\)

Equation of current : \(I=I_0\,sin\,\left(\omega t+\dfrac{\pi}{2}\right)\)

- Comparing the graphs, following observations can be taken:

\(\to\) When the graph of voltage starts from origin i.e., zero, then graph of current starts from its peak point \(I_0\) .

\(\to\) When the graph of voltage reaches at its peak point at 'a', then the graph of current reaches to zero at 'b'.

\(\to\) When the graph of voltage reaches to zero at 'c', then the graph of current reaches at its lowest point 'd'.

- From the observations, it can be concluded that for a sinusoidal applied voltage, the current in a capacitor is 90° out of phase i.e., 90° ahead of voltage across the capacitor.

Phasor is a vector which rotates counterclockwise about rhe origin with angular frequency \(\omega\).

- The length of the phasor is the maximum value of the quantity.

V_{0 }is the length of phasor.

- \(V_0\,sin\,\omega t\) is the projection on vertical axis. It is an instantaneous value.

- Phasor are useful to understand the phase difference between different quantities.
- If V is leading \(\text I\) by 90°.

- Consider an AC circuit in which only a capacitor is connected with AC voltage.
- When only capacitor is connected across battery, then such circuit is known as purely capacitive circuit.
- The capacitance of capacitor is C.
- The voltage across capacitor is \(V=V_0\,sin\,\omega t\).

\(\Rightarrow\) Charge on capacitor

Applying Kirchhoff's law,

\(V-\dfrac{q}{C}=0\)

\(V_0\,sin\,\omega t=\dfrac{q}{C}\)

\(q=CV_0\,sin\,\omega t\)

Here, q is the instantaneous charge on the capacitor.

\(\Rightarrow\) Current across capacitor

Differentiating the equation of charge on capacitor

\(\dfrac{dq}{dt}=i_C=\omega\, CV_0\,cos\,\omega t\)

\(\because\,cos\,\theta=sin\left(\dfrac{\pi}{2}+\theta\right)\)

\(\therefore \,i_C=\omega \,CV_0\,sin\left(\dfrac{\pi}{2}+\omega t\right)\)

- For a purely capacitive circuit,

peak current = \(\omega \,C\times\) peak voltage

\(I_0=\omega \,C\,V_0\)

\(I_0=\dfrac{V_0}{\left(\dfrac{1}{\omega \,C}\right)}\)

- This relationship between peak current and peak voltage looks much like Ohm's law for a resistor. \(\left(\dfrac{1}{\omega \,C}\right)\) behaves as resistance.

\(I=\dfrac{V}{R}\)

**Unit of \(\left(\dfrac{1}{\omega \,C}\right)\) **

- The unit of \(\dfrac{1}{\omega \,C}\), like those of resistance, is ohm.

**Capacitive reactance **

- The quantity \(\dfrac{1}{\omega \,C}\) is known as capacitive reactance. It is denoted by \(X_C\).

Mathematically, \(X_C=\dfrac{1}{\omega\, C}\)

**Dependence of \(X_C\) on frequency **

- Capacitive reactance is inversely proportional to frequency.

\(X_C\,\propto\,\dfrac{1}{\omega}\)

A decrease

B increase

C remain constant

D become half

- The square root of mean square current is called root-mean-square current or rms current.

\(i_{rms}=\sqrt {\overline{i^2}}\)

- Root mean square is a mathematical quantity.
- It is used to compare alternating current and direct current.
- Root mean square value of the alternating current is the direct current which when passed through a resistor for a given period of time would produce the same heat as that produced by alternating current when passed through the same resistor for the same time.

\(I_{rms}=\dfrac{V_{rms}}{X_C}\)

where, capacitive reactance, \(X_C=\dfrac{1}{\omega C}\)

- Consider an AC circuit in which only a capacitor is connected with AC voltage.
- When only capacitor is connected across battery, then such circuit is known as purely capacitive circuit.
- The capacitance of capacitor is C.
- The voltage across capacitor is \(V=V_0\,sin\,\omega t\).

\(\Rightarrow\) Charge on capacitor

Applying Kirchhoff's law,

\(V-\dfrac{q}{C}=0\)

\(V_0\,sin\,\omega t=\dfrac{q}{C}\)

\(q=CV_0\,sin\,\omega t\)

Here, q is the instantaneous charge on the capacitor.

\(\Rightarrow\) Current across capacitor

Differentiating the equation of charge on capacitor,

\(\dfrac{dq}{dt}=I_C=\omega\, CV_0\,cos\,\omega t\)

\(\because\,cos\,\theta=sin\left(\dfrac{\pi}{2}+\theta\right)\)

\(\therefore \,I_C=\omega\, C V_0\,sin\left(\dfrac{\pi}{2}+\omega t\right)\)