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.
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) [\(\because\; V_0=I_0\;\omega L \) ]
\(I=I_0\;sin\;\omega t\) ...(ii)
On comparing equation (i) and (ii) it can be concluded that
the voltage leads the current by \(\dfrac {\pi}{2}\) phase.
A \(3.3\,sin \left ( \omega t-\dfrac {\pi}{2} \right)\)
B \(4\,sin\;\omega t\)
C \(5\,sin\;\omega t\)
D \(6\,sin\;\omega t\)
The voltage source is given by \(V=V_0\;sin\;\omega t\).
The current is given by
\(I_L=I_0\,sin\left ( \omega t-\dfrac {\pi}{2}\right)\)
\(I_L=\dfrac {V_0}{\omega L}\,sin\left ( \omega t-\dfrac {\pi}{2}\right)\)
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
\(\because \;T=\dfrac {2\pi}{\omega}\)
\(\therefore \;I_L=-\dfrac {V_0}{\omega L}\,cos\, \left ( \dfrac {\omega ×2\pi}{\omega } \right)\)
\(I_L=\dfrac {-V_0}{\omega L}\)
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 :
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, V_{0 }is the length of phasor.
\(V_0\,sin\,\omega t\) is the projection on vertical axis. It is an instantaneous value.
Here, V is leading \(\text I\) by 90^{0}.
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.
\(I=\dfrac {V}{R}\)
Mathematically, \(X_L=\omega L\)
\(X_L\propto \omega\)
A increase
B decrease
C remain constant
D become half
\(i_{rms}=\sqrt {(\overline{i^2)}}\)
\(I_{rms}=\dfrac {V_{rms}}{X_L}\)
where Reactance, \(X_L=\omega L\)
Thus, \(I_{rms}=\dfrac {V_{rms}}{\omega L}\)
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) [\(\because\; V_0=I_0\;\omega L \) ]
\(I=I_0\;sin\;\omega t\) ...(ii)
On comparing equation (i) and (ii) it can be concluded that
the voltage is leading by \(\dfrac {\pi}{2}\) in phase with current.