Informative line

### Ac With Capacitor

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.

# Capacitor in an AC 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

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

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

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

#### In a purely capacitive circuit a voltage $$V=150\,sin\,\,120\,\pi t$$ is connected across a capacitor of capacitance $$C=8\,\mu F$$. Calculate the current as a function of time.

A $$0.9\,sin\,120\,\pi t$$

B $$0.45\,cos\,120\,\pi t$$

C $$10\,sin\,\,120\,\pi t$$

D $$15\,cos\,\,120\,\pi t$$

×

Voltage, $$V=150\,sin\,\,120\,\pi t$$

peak voltage, $$V_0=150\,V$$

capacitance of capacitor, $$C=8\,\mu F$$

angular frequency, $$\omega = 120\pi$$

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

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

Current, $$i_c=\omega\, CV_0\,cos\,120\,\pi t$$

$$i_c=120\,\pi×8×10^{-6}×150\,\,cos\,120\,\pi t$$

$$i_c=0.45\,cos\,120\,\pi t$$

### In a purely capacitive circuit a voltage $$V=150\,sin\,\,120\,\pi t$$ is connected across a capacitor of capacitance $$C=8\,\mu F$$. Calculate the current as a function of time.

A

$$0.9\,sin\,120\,\pi t$$

.

B

$$0.45\,cos\,120\,\pi t$$

C

$$10\,sin\,\,120\,\pi t$$

D

$$15\,cos\,\,120\,\pi t$$

Option B is Correct

# Phase Difference between Voltage & Current in Purely Capacitive Circuit

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

### Graph of voltage against time

• The voltage source is given by

$$V=V_0\,sin\,\omega t$$

### Graph of current against time

• The current is given by

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

### Steps to plot the graph of current against time

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,  $$t=T$$

$$\because \,T=\dfrac{2\pi}{\omega}$$

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

$$I_C=\omega \,CV_0$$

#### In a purely capacitive circuit, a voltage $$V=150\,sin\,120\,\pi t$$ is connected across a capacitor of capacitance $$C=8\,\mu F$$ . The graph of voltage against time is shown in figure. What will be the graph of current against time ?

A

B

C

D

×

Voltage, $$V=150\,sin\,120\,\pi t$$

Peak voltage, $$V_0=150\,V$$

Capacitance of capacitor, $$C=8\,\mu F$$

Angular frequency, $$\omega = 120\pi$$

The voltage across capacitor is  $$V=V_0\,sin\,\omega t$$

$$\Rightarrow$$ Charge on capacitor

Applying Kirchhoff's law-

$$V-\dfrac{q}{C}=0$$

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

$$q=C\,V_0\,sin\,\omega t$$

Here, q is the instantaneous charge on the capacitor

$$\Rightarrow$$ Current across capacitor

Differentiating the equation of charge on capacitor

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

Current, $$I_C=\omega\, CV_0\,cos\,120\,\pi t$$

$$I_C=120\pi×8×10^{-6}×150\,cos\,120\,\pi t$$

$$I_C=0.45\,cos\,120\,\pi t$$

At time, $$t=0$$

$$I_C=0.45\,cos\,120\,\pi(0)$$

$$I_C=0.45\,cos\,(0)$$

$$I_C=0.45\,A$$

At time, $$t=\dfrac{T}{4}=\dfrac{1}{4}\left(\dfrac{2\pi}{\omega}\right)=\dfrac{\pi}{2\omega}$$

$$I_C=0.45\,cos\,120\,\pi\left(\dfrac{\pi}{2×120\,\pi}\right)$$

$$I_C=0.45\,cos\left(\dfrac{\pi}{2}\right)$$

$$I_C=0$$

At time,  $$t=\dfrac{T}{2}=\dfrac{1}{2}\left(\dfrac{2\pi}{\omega}\right)=\dfrac{\pi}{\omega}$$

$$I_C=0.45\,cos\,120\,\pi\left(\dfrac{\pi}{120\,\pi}\right)$$

$$I_C=0.45\,cos\,\pi$$

$$I_C=-0.45\,A$$

At time,  $$t=\dfrac{3T}{4}=\dfrac{3}{4}\left(\dfrac{2\pi}{\omega}\right)=\dfrac{3}{2}\left(\dfrac{\pi}{\omega}\right)$$

$$I_C=0.45\,cos\,120\,\pi\left(\dfrac{3\,\pi}{2×120\,\pi}\right)$$

$$I_C=0.45\,cos\left(\dfrac{3\,\pi}{2}\right)$$

$$I_C=0$$

At time,  $$t=T=\dfrac{2\pi}{\omega}$$

$$I_C=0.45\,cos\,120\,\pi\left(\dfrac{2\,\pi}{120\,\pi}\right)$$

$$I_C=0.45\,A$$

On the basis of above steps, the graph will be obtained as-

Thus, option (B) is correct.

### In a purely capacitive circuit, a voltage $$V=150\,sin\,120\,\pi t$$ is connected across a capacitor of capacitance $$C=8\,\mu F$$ . The graph of voltage against time is shown in figure. What will be the graph of current against time ?

A
B
C
D

Option B is Correct

# Phasor Diagram for a Purely Capacitive Circuit

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

## Comparison between the graphs of voltage and current against time

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

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

### Length of Phasor

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

Vis the length of phasor.

• $$V_0\,sin\,\omega t$$ is the projection on vertical axis. It is an instantaneous value.

### Use of Phasor

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

### Phasor diagram of voltage and current

#### The voltage phasor for voltage $$V=150\,sin\,120\,\pi t$$ of a capacitor is shown in figure. If the peak current is $$I_0=0.45\,A$$, then choose the correct phasor of current.

A

B

C

D

×

Since, current phasor for a capacitor leads the voltage phasor by 90°.

Therefore, option (C) is correct.

Option (A) is incorrect because in this option, the current phasor lags behind the voltage phasor by 90°.

Option (B) is incorrect because the phase of current and voltage is same.

Option (D) is incorrect because the current phasor leads the voltage phasor by an angle more than 90°.

### The voltage phasor for voltage $$V=150\,sin\,120\,\pi t$$ of a capacitor is shown in figure. If the peak current is $$I_0=0.45\,A$$, then choose the correct phasor of current.

A
B
C
D

Option C is Correct

# Dependence of Capacitive Reactance on Frequency

• 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

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

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

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

## Comparison of Relation between $$I_0$$ & $$V_0$$ with Ohm's Law

• 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.

$$X_C\,\propto\,\dfrac{1}{\omega}$$

#### If frequency is increased in a purely capacitive circuit, as shown in figure, then the capacitive reactance will

A decrease

B increase

C remain constant

D become half

×

Since, $$X_C\,\propto\,\dfrac{1}{\omega\; }$$

Therefore, when frequency is increased, then capacitive reactance will decrease.

Hence, option (A) is correct.

### If frequency is increased in a purely capacitive circuit, as shown in figure, then the capacitive reactance will

A

decrease

.

B

increase

C

remain constant

D

become half

Option A is Correct

# Root Mean Square Value of Current in a Purely Capacitive AC Circuit

## Root Mean Square Value of Alternating Current

• 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 Value

• 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.

### Root Mean Square Value of Current in a Purely Capacitive AC Circuit

$$I_{rms}=\dfrac{V_{rms}}{X_C}$$

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

#### In a purely capacitive AC circuit, $$C=8\,\mu F$$ and rms voltage is $$V_{rms}=332\,V$$. Calculate the rms current in the circuit if frequency, $$f=60\,Hz$$.

A 2 A

B 1 A

C 3 A

D 4 A

×

rms voltage, $$V_{rms}=332\,V$$

capacitance of capacitor, $$C=8\,\mu F$$

frequency, $$f=60\,Hz$$

Capacitive reactance, $$X_C=\dfrac{1}{\omega \,C}$$

$$X_C=\dfrac{1}{2×3.14×60×8×10^{-6}}$$

$$X_C=332\,\Omega$$

rms current, $$I_{rms}=\dfrac{V_{rms}}{X_C}$$

$$I_{rms}=\dfrac{332}{332}$$

$$I_{rms}=1\,A$$

### In a purely capacitive AC circuit, $$C=8\,\mu F$$ and rms voltage is $$V_{rms}=332\,V$$. Calculate the rms current in the circuit if frequency, $$f=60\,Hz$$.

A

2 A

.

B

1 A

C

3 A

D

4 A

Option B is Correct

# Calculation of Current at a given Time when Voltage is given as a Function of Time

• 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,

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

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

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

#### In a purely capacitive circuit, the voltage source $$V=166\,sin\,120\,\pi t$$ is connected across capacitor of capacitance $$C=8\,\mu F$$. Find current at $$t=2\,sec$$.

A 0.1 A

B 0.5 A

C 0.6 A

D 1 A

×

Voltage, $$V=166\,sin\,120\,\pi t$$

Peak voltage, $$V_0=166\,V$$

Angular frequency, $$\omega=120\pi$$

Capacitance of capacitor, $$C=8\,\mu F$$

Time, $$t=2\,sec$$

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,

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

Current, $$I=I_0\,sin\left(\dfrac{\pi}{2}+120\pi t\right)$$

$$I=\omega\, CV_0\,cos\,120\,\pi t$$

$$I=120×3.14×8×10^{-6}×166\,cos\,120\,\pi (2)$$

$$I=0.5\,cos\,240\pi$$

$$I=0.5\,A$$     $$[\because\,cos\,240\pi=1]$$

### In a purely capacitive circuit, the voltage source $$V=166\,sin\,120\,\pi t$$ is connected across capacitor of capacitance $$C=8\,\mu F$$. Find current at $$t=2\,sec$$.

A

0.1 A

.

B

0.5 A

C

0.6 A

D

1 A

Option B is Correct