- #1
ZenSerpent
- 29
- 0
≪ mod note: thread moved from technical forum so template is missing ≫
Relate secondary current to primary voltage in a single-phase transformer having resistive load for non-ideal situation.
KVL, Voltage across resistor and inductor, ac voltage as a sine wave, reluctance, Faraday's law,other equations relevant to working principle of transformer.
Good day everyone! I would like to verify whether my attempt for relating secondary current to primary voltage in a single-phase transformer having resistive load is correct or not. Are there other losses that I have not taken into account? If so, feel free to critique my work. This is actually a work of my colleagues and I for our research. So here's our go at it:Using KVL for the equivalent circuit in the primary side gives,[tex] v_1=v_{R_1}+v_{L_1} [/tex]Substituting their corresponding values gives
[tex]v_{1max}{\mathrm{sin} \omega t\ }=i_1R_1+L_1\frac{di_1}{dt}[/tex]Solving DE for [itex] i_1 [/itex] gives
[tex] i_1=\frac{v_{1max}}{R^2_1+{{\omega }^2L}^2_1}\left(R_1{\mathrm{sin} \omega t\ }-\omega L_1{\mathrm{cos} \omega t\ }\right)\ [/tex]Combining the sine wave and cosine wave and simplifying gives
[tex] i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]Now since [itex] N_1i_1=\mathcal{R}\mathrm{\Phi }\ [/itex], solving for [itex] {\Phi } [/itex] gives
[itex] \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ [/itex]. This is the flux that flows to the secondary circuit after applying reluctance of the core.
Substituting [itex] i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/itex] to [itex] \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ [/itex] gives[tex] \mathrm{\Phi }=\frac{N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]
To determine the flux linkage in the secondary winding, we multiply N2 to both sides, giving[tex] N_2\mathrm{\Phi }=\frac{N_2N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]Taking derivative wrt t gives the voltage across the secondary circuit. That is,[tex] v_2=N_2\frac{d\mathrm{\Phi }}{dt}=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]Using KVL on the secondary circuit, we get[tex] v_2=v_{R_2}+v_{R_L}+v_{L_2}\ [/tex]Substituting their corresponding values give,[tex] \frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2R_2+i_2R_L+L_2\frac{di_2}{dt}\ [/tex][tex] \frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2(R_2+R_L)+L_2\frac{di_2}{dt}\ [/tex]Solving DE for [itex] i_2 [/itex] gives,
[itex] i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\left[{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2\right]}\left[\left(R_2+R_L\right){\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }+\omega L_2{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\right]\ [/itex]Combining the sine and cosine waves gives and simplifying gives,
[tex] i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\sqrt{{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2}}{\mathrm{sin} \left\{\omega t+{\mathrm{arctan} \left[\frac{R_1\left(R_2+R_L\right)-{\omega }^2L_1L_2}{R_1\omega L_2+\left(R_2+R_L\right)\omega L_1}\right]\ }\right\}\ }\ [/tex]
Critiquing this will be of great help for our group's thesis. Thank you so much. :)
Homework Statement
Relate secondary current to primary voltage in a single-phase transformer having resistive load for non-ideal situation.
Homework Equations
KVL, Voltage across resistor and inductor, ac voltage as a sine wave, reluctance, Faraday's law,other equations relevant to working principle of transformer.
The Attempt at a Solution
Good day everyone! I would like to verify whether my attempt for relating secondary current to primary voltage in a single-phase transformer having resistive load is correct or not. Are there other losses that I have not taken into account? If so, feel free to critique my work. This is actually a work of my colleagues and I for our research. So here's our go at it:Using KVL for the equivalent circuit in the primary side gives,[tex] v_1=v_{R_1}+v_{L_1} [/tex]Substituting their corresponding values gives
[tex]v_{1max}{\mathrm{sin} \omega t\ }=i_1R_1+L_1\frac{di_1}{dt}[/tex]Solving DE for [itex] i_1 [/itex] gives
[tex] i_1=\frac{v_{1max}}{R^2_1+{{\omega }^2L}^2_1}\left(R_1{\mathrm{sin} \omega t\ }-\omega L_1{\mathrm{cos} \omega t\ }\right)\ [/tex]Combining the sine wave and cosine wave and simplifying gives
[tex] i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]Now since [itex] N_1i_1=\mathcal{R}\mathrm{\Phi }\ [/itex], solving for [itex] {\Phi } [/itex] gives
[itex] \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ [/itex]. This is the flux that flows to the secondary circuit after applying reluctance of the core.
Substituting [itex] i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/itex] to [itex] \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ [/itex] gives[tex] \mathrm{\Phi }=\frac{N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]
To determine the flux linkage in the secondary winding, we multiply N2 to both sides, giving[tex] N_2\mathrm{\Phi }=\frac{N_2N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]Taking derivative wrt t gives the voltage across the secondary circuit. That is,[tex] v_2=N_2\frac{d\mathrm{\Phi }}{dt}=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]Using KVL on the secondary circuit, we get[tex] v_2=v_{R_2}+v_{R_L}+v_{L_2}\ [/tex]Substituting their corresponding values give,[tex] \frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2R_2+i_2R_L+L_2\frac{di_2}{dt}\ [/tex][tex] \frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2(R_2+R_L)+L_2\frac{di_2}{dt}\ [/tex]Solving DE for [itex] i_2 [/itex] gives,
[itex] i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\left[{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2\right]}\left[\left(R_2+R_L\right){\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }+\omega L_2{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\right]\ [/itex]Combining the sine and cosine waves gives and simplifying gives,
[tex] i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\sqrt{{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2}}{\mathrm{sin} \left\{\omega t+{\mathrm{arctan} \left[\frac{R_1\left(R_2+R_L\right)-{\omega }^2L_1L_2}{R_1\omega L_2+\left(R_2+R_L\right)\omega L_1}\right]\ }\right\}\ }\ [/tex]
Critiquing this will be of great help for our group's thesis. Thank you so much. :)
Last edited: