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Ac With Inductance

Practice inductive reactance of purely inductive circuit equation, learn purely inductive circuit, an inductor of inductance, is connected across a voltage at frequency. Calculate the value of current as a function of time.

Inductor in an AC Circuit

  • Consider an AC circuit in which only an inductor is connected with AC voltage.
  • When only inductor is connected across battery, then such circuit is known as purely inductive circuit.
  • The self inductance of inductor is L.
  • The resistance of inductor is zero.
  • The current flowing in circuit is \(I=I_0\;sin\,\omega t\).

Voltage Across the Inductor

Applying Kirchhoff's Law on the circuit,

\(V-L\dfrac {dI}{dt}=0\)

 \(\because\;\;\;\dfrac {dI}{dt}=\dfrac {d}{dt}(I_0\;sin\,\omega t)\)

\(\dfrac {dI}{dt}=I_0\;\omega \,cos\,\omega t\)

 \(\therefore \;\;V=L\;I_0\,\omega\;cos\;\omega t\)

\(V=I_0\;\omega L\;sin\left (\omega t+\dfrac {\pi}{2}\right)\)

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

\(V=V_0\;sin\,\left (\omega t+\dfrac {\pi}{2}\right)\) ...(i)              [\(\because\; V_0=I_0\;\omega L \) ]

Current Across the Inductor

\(I=I_0\;sin\;\omega t\) ...(ii)

Relation between Current and Voltage Phasor

On comparing equation (i) and (ii) it can be concluded that

the voltage leads the current by \(\dfrac {\pi}{2}\) phase.

 

Illustration Questions

In a purely inductive circuit, an inductor of inductance \(L = 200\, mH\), is connected across a voltage of  \(V=210\,sin\;\omega t\) at frequency of  \(f = 50\, Hz\) . Calculate the value of current as a function of time.

A  \(3.3\,sin \left ( \omega t-\dfrac {\pi}{2} \right)\)

B  \(4\,sin\;\omega t\)

C  \(5\,sin\;\omega t\)

D  \(6\,sin\;\omega t\)

×

Inductance of inductor, \(L = 200\, mH\)

Voltage, \(V=210\,sin\;\omega t\)

Frequency,\( f = 50\, Hz\)

Amplitude of current, \(I_0=\dfrac {V_0}{\omega L}\)

\(I_0=\dfrac {210}{2×\pi × 50×200×10^{-3}}\)

\(I_0=3.3A\)

The voltage leads the current by \(\dfrac {\pi}{2}\) phase.

Current, \(\text I\) =  \(3.3\,sin \left ( \omega t-\dfrac {\pi}{2} \right)\)

 

So, option (A) is correct.

In a purely inductive circuit, an inductor of inductance \(L = 200\, mH\), is connected across a voltage of  \(V=210\,sin\;\omega t\) at frequency of  \(f = 50\, Hz\) . Calculate the value of current as a function of time.

A

 \(3.3\,sin \left ( \omega t-\dfrac {\pi}{2} \right)\)

.

B

 \(4\,sin\;\omega t\)

C

 \(5\,sin\;\omega t\)

D

 \(6\,sin\;\omega t\)

Option A is Correct

Graph of Voltage and Current with Time in Purely Inductive Circuit

  • Consider a purely inductive 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 

\(I_L=I_0\,sin\left ( \omega t-\dfrac {\pi}{2}\right)\)

\(I_L=\dfrac {V_0}{\omega L}\,sin\left ( \omega t-\dfrac {\pi}{2}\right)\)

Steps to Plot the Graph of Current against Time

Step 1 : Convert the equation of current in terms of cosine.

\(I_L=-\dfrac {V_0}{\omega L}\,cos\, \omega t\)

Step 2 : At time t = 0

\(I_L=-\dfrac {V_0}{\omega L}\,cos\,(0)\)

\(I_L=-\dfrac {V_0}{\omega L}\)

Step 3 : At time \(t=\dfrac {T}{4}\)

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

\(\therefore \;I_L=-\dfrac {V_0}{\omega L}\,cos\, \left ( \dfrac {\omega \pi}{2\omega } \right)\)

 \(I_L=0 \)

Step 4 : At time \(t=\dfrac {T}{2}\)

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

 \(\therefore\; \,I_L=-\dfrac {V_0}{\omega L}\,cos\, \left ( \dfrac {\omega \pi}{\omega } \right)\\=+\dfrac {V_0}{\omega L}\)

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

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

 \(\therefore \;I_L=-\dfrac {V_0}{\omega L}\,cos\, \left ( \dfrac {\omega ×3\pi}{2\omega } \right)\)

\(I_L=-\dfrac {V_0}{\omega L}\,cos\, \left ( \dfrac {3\pi}{2} \right)\)

\(I_L=0\)

Step 6 : At time t = T

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

 \(\therefore \;I_L=-\dfrac {V_0}{\omega L}\,cos\, \left ( \dfrac {\omega ×2\pi}{\omega } \right)\)

 \(I_L=\dfrac {-V_0}{\omega L}\)

Illustration Questions

In a purely inductive circuit, an inductor of inductance L = 200 mH, is connected across a voltage of V = 210 \(sin\;\omega t\) at frequency of f = 50 Hz. The graph of voltage is shown in figure. What will be the graph of current against time?

A

B

C

D

×

Amplitude of current, 

\(I_0=\dfrac {V_0}{\omega L}\)

\(I_0=\dfrac {210}{2×3.14×50×200×10^{-3}}\)

\(I_0\)\(3.3\,A\)\(\)

For a sinusoidal applied voltage, the current in an inductor always lags behind the voltage across the inductor by 90°.

Thus, current will be

\(I=\)\(3.3\,sin\left ( \omega t-\dfrac {\pi}{2}\right)\)

\(\Rightarrow I=\)\(3.3\,cos\,\omega t\)

At time, t = 0  

\(I_L=\) –3.3 \(cos\;\omega\) (0) = – 3.3 \(A\)

 

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

 \(I_L=\) –3.3 \(cos\;\left (\dfrac {\omega\;\pi}{2\omega}\right)\) = 0 \(A\) 

 

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

 \(I_L=\) –3.3 \(cos\;\left (\dfrac {\omega\;\pi}{\omega}\right)\) = 3.3 \(A\)

 

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

 \(I_L=\) –3.3 \(cos\;\left (\dfrac {\omega\;3\pi}{2\omega}\right)\) = 0 \(A\)

 

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

\(I_L=\) –3.3 \(cos\;\left (\dfrac {\omega\;2\pi}{\omega}\right)\) = –3.3 \(A\)

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

image

Thus, option (A) is correct.

In a purely inductive circuit, an inductor of inductance L = 200 mH, is connected across a voltage of V = 210 \(sin\;\omega t\) at frequency of f = 50 Hz. The graph of voltage is shown in figure. What will be the graph of current against time?

image
A image
B image
C image
D image

Option A is Correct

Phase Difference between Current and Voltage in a Purely Inductive Circuit

  • Consider a purely inductive circuit as shown in figure.

Comparison between the Graphs of Voltage and Current against Time

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

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

where

\(I_0=\dfrac {V_0}{\omega L}\)

Comparing the graphs, following observations can be taken :

  • The graph of voltage starts from origin and graph of current starts from –\(I_0\).
  • When the graph of voltage reaches at its peak point at 'a' then the graph of current reaches at zero at 'b'.
  • When the graph of voltage reaches at zero at 'c', then the graph of current reaches at its peak point at 'd'.

From the observations, it can be concluded that for a sinusoidal applied voltage, the current in an inductor always lags behind the voltage across the inductor by 90°.

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

  • Length of Phasor : The length of the phasor is the maximum value of the quantity.

          Here, 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.

           Here, V is leading \(\text I\) by 900.

 

Phasor Diagram of Voltage and Current

  • In an inductor, current always lags behind the voltage by 90°.

Illustration Questions

The current phasor for current  \(I=I_0\,sin\;\omega t\) of an inductor is shown in figure. If the peak voltage is \(V_0\), then choose the correct phasor of voltage.

A

B

C

D

×

Since, the voltage phasor for an inductor leads the current phasor by 90°.

Therefore, option (C) is correct.

image

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

image

Option (B) is incorrect, because the voltage phasor lags behind the current phasor by 90°.

image

Option (D) is incorrect, because the voltage phasor leads the current phasor by \((90°+\omega t)\).

image

The current phasor for current  \(I=I_0\,sin\;\omega t\) of an inductor is shown in figure. If the peak voltage is \(V_0\), then choose the correct phasor of voltage.

image
A image
B image
C image
D image

Option C is Correct

Inductive Reactance of Purely Inductive Circuit

  • For a purely inductive circuit,

      Peak current = \(\dfrac {\text {Peak voltage}}{\omega L}\)

                  \(I_0=\dfrac {V_0}{\omega L}\)

This relationship between peak current and  peak voltage looks much like Ohm's law for a resistor. \(\omega L\) behaves as resistance.

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

Unit of \(\omega L\)

  • The unit of \(\omega L\), like those of resistance, is Ohm.

Inductive Reactance

  • The quantity \(\omega L\) is known as inductive reactance. It is denoted by \(X_L\).

               Mathematically, \(X_L=\omega L\)

Work of Inductive Reactance

  • It opposes the flow of charge.

Dependence of \(X_L\) on Frequency

  • Inductive reactance is directly proportional to frequency.

\(X_L\propto \omega\)

 

Illustration Questions

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

A increase

B decrease

C remain constant

D become half

×

Since, \(X_L\propto \omega\)

Therefore, when frequency increases, then reactance also increases.

Option (A) is correct.

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

image
A

increase

.

B

decrease

C

remain constant

D

become half

Option A is Correct

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 value 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 Inductive AC Circuit

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

where  Reactance, \(X_L=\omega L\)

Thus, \(I_{rms}=\dfrac {V_{rms}}{\omega L}\)       

Illustration Questions

In a purely inductive AC circuit, L = 25 mH and rms voltage is Vrms = 150 V. Calculate the rms current in the circuit if frequency, f = 60 Hz.

A 16.4 A

B 15.9 A

C 20 A

D 17 A

×

Inductance of inductor, L = 25 mH

rms voltage, Vrms = 150 V

Frequency, f = 60 Hz

rms current, \(I_{rms}=\dfrac {V_{rms}}{\omega L}\)

\(I_{rms}=\dfrac {150}{2×3.14×60×25×10^{-3}}\)

\(I_{rms}=\) 15.9 A

In a purely inductive AC circuit, L = 25 mH and rms voltage is Vrms = 150 V. Calculate the rms current in the circuit if frequency, f = 60 Hz.

A

16.4 A

.

B

15.9 A

C

20 A

D

17 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 an inductor is connected with AC voltage.
  • When only inductor is connected across battery, then such circuit is known as purely inductive circuit.
  • The self inductance of inductor is L.
  • The resistance of inductor is zero.
  • The current flowing in circuit is \(I=I_0\;sin\,\omega t\).

Voltage Across the Inductor

Applying Kirchhoff's Law on the circuit,

\(V-L\dfrac {dI}{dt}=0\)

 \(\because\;\;\;\dfrac {dI}{dt}=\dfrac {d}{dt}(I_0\;sin\,\omega t)\)

\(\dfrac {dI}{dt}=I_0\;\omega \,cos\,\omega t\)

 \(\therefore \;\;V=L\;I_0\,\omega\;cos\;\omega t\)

\(V=I_0\;\omega L\;sin\left (\omega t+\dfrac {\pi}{2}\right)\)

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

\(V=V_0\;sin\,\left (\omega t+\dfrac {\pi}{2}\right)\) ...(i)              [\(\because\; V_0=I_0\;\omega L \) ]

Current Across the Inductor

\(I=I_0\;sin\;\omega t\) ...(ii)

Relation between Current and Voltage Phasor

On comparing equation (i) and (ii) it can be concluded that

the voltage is leading by \(\dfrac {\pi}{2}\) in phase with current.

Illustration Questions

In a purely inductive circuit, V = 210 sin 120 \(\pi t\)  and inductance of inductor is L = 25 mH. Calculate current in the circuit at time t = 2 second.

A – 22.29 A

B 40 A

C 20 A

D 50 A

×

Peak voltage, \(V_0=\) 210 V

Angular frequency,  \(\omega=\) 120 \(\pi\,rad/sec\) 

Inductance of inductor, L = 25 mH

Time, t = 2 sec

Voltage, V = 210 sin 120 \(\pi t\)

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

\(I=-\dfrac {V_0}{\omega L}\;cos\, 120\;\pi\,t\)

\(I=\dfrac {-210}{120×3.14 × 25×10^{-3}}\, cos\; 120\;\pi (2)\)

\(I = – 22.29\; cos\; 240 \pi\)             [ \(\therefore cos\, n \pi=1\) , if n = even ]

\(I = – 22.29 \,A\)    

In a purely inductive circuit, V = 210 sin 120 \(\pi t\)  and inductance of inductor is L = 25 mH. Calculate current in the circuit at time t = 2 second.

A

– 22.29 A

.

B

40 A

C

20 A

D

50 A

Option A is Correct

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