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Merge pull request #22 from virtual-labs/dev
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jnipun1994 committed Dec 4, 2023
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74 changes: 35 additions & 39 deletions experiment/theory.md
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Expand Up @@ -52,15 +52,11 @@ current and no voltage is applied to the load R.<br>

<b>Analysis</b> : Average load voltage is given by<br>

<center>

$$ V_dc = \frac {1}{2\pi} [\int_\alpha^\pi V_max, sin , \omega t , d(\omega t)+\int_\pi^{2\pi}(0)d(\omega t)] $$

</center><br>

<center>

$$ V_dc = \frac {1}{2\pi} [\int_\alpha^\pi V_{max}, sin , \omega t , d(\omega t)+\int_\pi^{2\pi}(0)d(\omega t)] $$
$$ V_dc = \frac {1}{2\pi} [\int_\alpha^\pi V_{max} sin \omega t d(\omega t)+\int_\pi^{2\pi}(0)d(\omega t)].......(1) $$

</center>

Expand All @@ -70,7 +66,7 @@ for α = 0 and α = π or 180°. The maximum output voltage is obtained when α

<center>

$V_{dc \;max} =\frac {V_{max}}{\pi} .......(2)$
$$ V_{dc \;max} =\frac {V_{max}}{\pi} .......(2) $$

</center>

Expand All @@ -86,31 +82,31 @@ The rms value of load voltage is given by<br>

<center>

$V_{L rms}=\frac {V_{max}}{2 \sqrt{\pi} } (\pi -\alpha + \frac {1}{2}sin \,2 \alpha)^\frac {1}{2}.......(4)$
$$V_{L rms}=\frac {V_{max}}{2 \sqrt{\pi} } (\pi -\alpha + \frac {1}{2}sin \,2 \alpha)^\frac {1}{2}.......(4)$$

</center>

For firing angle α =0<br>

<center>

$V_{L \, rms} = \frac {V_{max}}{2}.......(5)$
$$V_{L \, rms} = \frac {V_{max}}{2}.......(5)$$

</center>

The rms value of load current<br>

<center>

$I_{L \, rms} = \frac {V_{L \, rms}}{R} = \frac {V_{L \, rms}}{2 R \sqrt{\pi}}(\pi -\alpha + \frac {1}{2}sin \,2 \alpha)^\frac {1}{2}......(6)$
$$I_{L \, rms} = \frac {V_{L \, rms}}{R} = \frac {V_{L \, rms}}{2 R \sqrt{\pi}}(\pi -\alpha + \frac {1}{2}sin \,2 \alpha)^\frac {1}{2}......(6)$$

</center>

Average output power is given by <br>

<center>

$P_{dc} = V_{dc}.I_{dc} = \frac {V_{max}^2}{4 \pi^2 R}(1+ cos \alpha)^ \frac {1}{2}.......(7)$
$$P_{dc} = V_{dc}.I_{dc} = \frac {V_{max}^2}{4 \pi^2 R}(1+ cos \alpha)^ \frac {1}{2}.......(7)$$

</center>

Expand Down Expand Up @@ -157,7 +153,7 @@ The Voltage equation is :<br>

<center>

$V_{m}\,sin \, \omega t = Ri_0+L \frac {dI_0}{dt}$
$$V_{m}\,sin \, \omega t = Ri_0+L \frac {dI_0}{dt}$$

</center>

Expand All @@ -166,45 +162,45 @@ transient component i<sub>t</sub>. Here i<sub>s</sub> is given by<br>

<center>

$i_{s} = \frac {V_m}{\sqrt{ (R^2+X^2)}}sin(\omega t- \phi)$
$$i_{s} = \frac {V_m}{\sqrt{ (R^2+X^2)}}sin(\omega t- \phi)$$

</center>

The transient component i<sub>t</sub> can be obtained from force free equation<br>

<center>

$Ri_t + L\frac{di_t}{dt} = 0$
$$Ri_t + L\frac{di_t}{dt} = 0$$

</center>

Its solution gives,<br>

<center>

$i_t = A e^{-(\frac{R}{L})t}$
$$i_t = A e^{-(\frac{R}{L})t}$$

</center>

<center>

$i_o = i_s + i_t =\frac {V_m}{Z} sin(\omega t-\phi) + Ae^{-(\frac {R}{L})t} .........(8)$
$$i_o = i_s + i_t =\frac {V_m}{Z} sin(\omega t-\phi) + Ae^{-(\frac {R}{L})t} .........(8)$$

</center>

Constant A can be obtained from the boundary condition at &omega;t = &alpha;. At time $t = \frac{\alpha}{\omega}$, $i_o = 0$ . Thus from equation (8)
Constant A can be obtained from the boundary condition at &omega;t = &alpha;. At time $$t = \frac{\alpha}{\omega}$$, $$i_o = 0$$ . Thus from equation (8)

<center>

$A = -\frac {V_m}{Z} sin(\alpha - \phi )e^{\frac {R \alpha}{L \omega}}$
$$A = -\frac {V_m}{Z} sin(\alpha - \phi )e^{\frac {R \alpha}{L \omega}}$$

</center>

Substitution of A in Eq. (8) gives <br>

<center>

$i_o = \frac {V_m}{Z} sin(\omega t- \phi) -\frac {V_m}{Z}sin(\alpha -\phi ) exp \{-\frac{R}{\omega L}(\omega t-\alpha )\}.......(9)$
$$i_o = \frac {V_m}{Z} sin(\omega t- \phi) -\frac {V_m}{Z}sin(\alpha -\phi ) exp \{-\frac{R}{\omega L}(\omega t-\alpha )\}.......(9)$$

</center>

Expand All @@ -214,29 +210,29 @@ The average voltage V<sub>o</sub> is given by<br>

<center>

$V_o = \frac {1}{2 \pi} \int_\alpha ^\beta V_m sin\, \omega t \, d(\omega t) =\frac {V_m}{2 \pi}(cos \alpha -cos \beta).......(10)$
$$V_o = \frac {1}{2 \pi} \int_\alpha ^\beta V_m sin\, \omega t \, d(\omega t) =\frac {V_m}{2 \pi}(cos \alpha -cos \beta).......(10)$$

</center>

The average load currrent I<sub>o</sub> is given by <br>

<center>

$I_o = \frac {V_m}{2 \pi R}(cos \alpha -cos \beta)$
$$I_o = \frac {V_m}{2 \pi R}(cos \alpha -cos \beta)$$

</center>

Rms Load voltage is<br>

<center>

$V_{or} = [ \frac {1}{2 \pi } \int_\alpha ^\beta V_m^2 \, sin^2 \, \omega t.d(\omega t)]^{\frac {1}{2}}$
$$V_{or} = [ \frac {1}{2 \pi } \int_\alpha ^\beta V_m^2 \, sin^2 \, \omega t.d(\omega t)]^{\frac {1}{2}}$$

</center>

<center><br>

$V_{or} = \frac {V_m}{2 \pi} \sqrt {[(\beta - \alpha)- \frac {1}{2}(sin 2 \beta - sin 2 \alpha )]}........(11)$
$$V_{or} = \frac {V_m}{2 \pi} \sqrt {[(\beta - \alpha)- \frac {1}{2}(sin 2 \beta - sin 2 \alpha )]}........(11)$$

</center>

Expand All @@ -250,7 +246,7 @@ the load may be due to a battery or a DC motor. The minimum value of firing angl

<center>

$\theta_1 = sin^{-1}(\frac{E}{V_m}).......(12)$
$$\theta_1 = sin^{-1}(\frac{E}{V_m}).......(12)$$

</center>

Expand All @@ -263,7 +259,7 @@ curve, KVL gives the voltage differential equation as <br>

<center>

$V_m \, sin \omega t = Ri_o + L\frac{di_o}{dt} + E.........(13)$
$$V_m \, sin \omega t = Ri_o + L\frac{di_o}{dt} + E.........(13)$$

</center><br>

Expand All @@ -275,66 +271,66 @@ i<sub>s1</sub> is the steady state current due to ac source voltage acting alone

<center>

$i_{s1} = \frac{V_m}{Z} sin(\omega t-\phi)$
$$i_{s1} = \frac{V_m}{Z} sin(\omega t-\phi)$$

</center>

If only E were present, then steady state current i<sub>s2</sub> would be given by<br>

<center>

$i_{s2} = -(\frac{E}{R})$
$$i_{s2} = -(\frac{E}{R})$$

</center><br>

The transient current i<sub>t</sub> is given by $i_t = Ae^{-(\frac{R}{L})^t}$<br>
The transient current i<sub>t</sub> is given by $$i_t = Ae^{-(\frac{R}{L})^t}$$<br>

At &omega;t = &alpha;, $i_o = 0$, i.e., at $t = \frac{\alpha}{\omega}$, $i_o = 0$.
At &omega;t = &alpha;, $$i_o = 0$$, i.e., at $$t = \frac{\alpha}{\omega}$$, $$i_o = 0$$.

<center>

$A = [\frac {E}{R}-\frac {V_m}{Z} sin(\alpha - \phi )]e^{-\frac{Ra}{L \omega}}$
$$A = [\frac {E}{R}-\frac {V_m}{Z} sin(\alpha - \phi )]e^{-\frac{Ra}{L \omega}}$$

</center>

<center>

$i_o =\frac {V_m}{Z}[sin(\omega t- \phi )-sin( \alpha - \phi)exp\{-\frac {R}{\omega L}(\omega t- \alpha )\}]-\frac {E}{R}[1-exp\{-\frac{R}{ \omega L}(\omega t- \alpha)\}].......(14)$
$$i_o =\frac {V_m}{Z}[sin(\omega t- \phi )-sin( \alpha - \phi)exp\{-\frac {R}{\omega L}(\omega t- \alpha )\}]-\frac {E}{R}[1-exp\{-\frac{R}{ \omega L}(\omega t- \alpha)\}].......(14)$$

</center>

Equation 11 is applicable for &alpha; &le; &omega;t &ge; &beta;. The extinction angle &beta; depends upon load emf E, firing angle &alpha; and the load impedence angle &phi; the average load current $I_o$ is given by<br>
Equation 11 is applicable for &alpha; &le; &omega;t &ge; &beta;. The extinction angle &beta; depends upon load emf E, firing angle &alpha; and the load impedence angle &phi; the average load current $$I_o$$ is given by<br>

<center>

$I_o = \frac{1}{2 \pi R}[V_m(cos \alpha -cos \beta )-E(\beta-\alpha)]..........(15)$
$$I_o = \frac{1}{2 \pi R}[V_m(cos \alpha -cos \beta )-E(\beta-\alpha)]..........(15)$$

</center>

Average load voltage V<sub>o</sub> is given by<br>

<center>

$ V_o= E+I_oR=E+ \frac {1}{2 \pi}[2V_m sin(\alpha + \frac{\gamma}{2})sin(\frac {\gamma}{2})-\gamma.E]$
$$ V_o= E+I_oR=E+ \frac {1}{2 \pi}[2V_m sin(\alpha + \frac{\gamma}{2})sin(\frac {\gamma}{2})-\gamma.E]$$

</center><br>

<center>

$ V_o=E(1-\frac {\gamma} {2 \pi})+\frac {V_m} {\pi} sin(\alpha +\frac {\gamma}{2})sin(\frac {\gamma}{2})..........(16)$
$$ V_o=E(1-\frac {\gamma} {2 \pi})+\frac {V_m} {\pi} sin(\alpha +\frac {\gamma}{2})sin(\frac {\gamma}{2})..........(16)$$

</center><br>

In case &beta; is made equal to (&gamma;+&alpha;) in the above expression, Equation (15) can be obtained.
If load inductance L is zero in Fig. 5, then extinction angle &beta; = $\theta_2$ and &gamma; = &beta; - &alpha;
If load inductance L is zero in Fig. 5, then extinction angle &beta; = $$\theta_2$$ and &gamma; = &beta; - &alpha;
= &theta;<sub>2</sub> - &alpha; but &theta;<sub>2</sub> = &pi; - &theta;<sub>1</sub>.<br>

This gives &beta; = &theta;<sub>2</sub> = &pi; - &theta;<sub>1</sub> and &gamma; = &pi; - &theta;<sub>1</sub> - &alpha;.
Putting this in Equation (15), we get average load current I<sub>o</sub> as under :<br>

<center>

$I_o = \frac {1}{ 2 \pi R}[V_m (cos \alpha +cos \theta_1)-E( \pi - \theta_1- \alpha)].......(17)$
$$I_o = \frac {1}{ 2 \pi R}[V_m (cos \alpha +cos \theta_1)-E( \pi - \theta_1- \alpha)].......(17)$$

</center><br>

Expand All @@ -343,17 +339,17 @@ as under:<br>

<center>

$I_{or}=[\frac {1}{2 \pi R^2} \int_\alpha ^{\pi-\alpha}\{(V_m sin \omega t-E)^2.d(\omega t)\}]^\frac {1}{2}.........(18)$
$$I_{or}=[\frac {1}{2 \pi R^2} \int_\alpha ^{\pi-\alpha}\{(V_m sin \omega t-E)^2.d(\omega t)\}]^\frac {1}{2}.........(18)$$

</center><br>

Power delivered to load<br>

<center>$ P=I^2R+I_oE.......(19)$</center><br>
<center>$$ P=I^2R+I_oE.......(19)$$</center><br>

<center>

Supply power factor =$I_or^2R+I_o \frac {E}{V_sI_{or}}$
Supply power factor =$$I_or^2R+I_o \frac {E}{V_sI_{or}}$$

</center>

Expand Down

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