Hi,
I have a circuit which I want to analyse using phasor at its natural frequency.
The circuit consists of a parallel inductor-capacitor [latex](L || C_1)[/latex] in series with a parallel resistor-capacitor [latex](R || C_2)[/latex] which is connected to an AC voltage source. So we have parallel inductor-capacitor impedence, call it [latex] Z_1 [/latex], and parallel resistor-capacitor impendence, call it [latex] Z_2 [/latex].
Therefore their impedences are:
[latex]Z_1 = \frac{1}{\frac{1}{j \omega L} - \frac{\omega C_1}{j}} = j\frac{\omega L}{1 - \omega^2 L C_1}[/latex]
and,
[latex]Z_2 = \frac{1}{\frac{1}{R} - \frac{\omega C_2}{j}} = \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2}.[/latex]
Since these impendences are in series the total impedence of the circuit is:
[latex]Z_1 + Z_2= j\frac{\omega L}{1 - \omega^2 L C_1} + \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = \frac{\frac{1}{R}}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} + j \left[ \frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} \right].[/latex]
The natural frequency of this circuit occurs when the reactance of the series is zero so,
[latex]\frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = 0.[/latex]
Rearranging this we get,
[latex]\omega_o = \sqrt{\frac{C_2 -\frac{L}{R^2}}{LC_2^2 + L C_1 C_2}}.[/latex]
Let the voltage source be denoted by [latex]V_s[/latex]. At natural frequency, [latex]\omega_o[/latex], the voltage across the impedence [latex]Z_1[/latex] is,
[latex]V_1 = \frac{Z_1}{Z_1 + Z_2} \times V_s = j \frac{\frac{\omega_o L}{1 - \omega_o^2 L C_1}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \times V_s[/latex]
and the voltage across the impedence [latex]Z_2[/latex] is,
[latex]V_2 = \frac{Z_2}{Z_1 + Z_2} \times V_s = \left[ 1 - j \left( \frac{ \frac{\omega_o C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \right) \right] \times V_s = \left[ 1 - j \left( \frac{\omega_o C_2}{R \left[ \left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2 \right]^2} \right) \right] \times V_s[/latex]
Okay, now let [latex]V_s = 12, \ L = 0.1mH, \ C_1 = 100nF, \ C_2 = 80nF, \ R = 100 \Omega[/latex]. Using the formulae above I got,
[latex]\omega_o = 220479.275[/latex]Hz, [latex]V_1 = 12 \times (j 1.04364 \times 10^{3})[/latex], and [latex]V_2 = 12 \times (1 - j 1.04364 \times 10^{3})[/latex].
These mean [latex]|V_1| = 12.52368 \times 10^3[/latex] and [latex]|V_2| = 12.52368 \times 10^3[/latex]. But my simulation using SPICE, I got both [latex]|V_1|[/latex] and [latex]|V_2|[/latex] roughly around 12 volts only! Can someone tell me where I am wrong?
I have a circuit which I want to analyse using phasor at its natural frequency.
The circuit consists of a parallel inductor-capacitor [latex](L || C_1)[/latex] in series with a parallel resistor-capacitor [latex](R || C_2)[/latex] which is connected to an AC voltage source. So we have parallel inductor-capacitor impedence, call it [latex] Z_1 [/latex], and parallel resistor-capacitor impendence, call it [latex] Z_2 [/latex].
Therefore their impedences are:
[latex]Z_1 = \frac{1}{\frac{1}{j \omega L} - \frac{\omega C_1}{j}} = j\frac{\omega L}{1 - \omega^2 L C_1}[/latex]
and,
[latex]Z_2 = \frac{1}{\frac{1}{R} - \frac{\omega C_2}{j}} = \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2}.[/latex]
Since these impendences are in series the total impedence of the circuit is:
[latex]Z_1 + Z_2= j\frac{\omega L}{1 - \omega^2 L C_1} + \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = \frac{\frac{1}{R}}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} + j \left[ \frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} \right].[/latex]
The natural frequency of this circuit occurs when the reactance of the series is zero so,
[latex]\frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = 0.[/latex]
Rearranging this we get,
[latex]\omega_o = \sqrt{\frac{C_2 -\frac{L}{R^2}}{LC_2^2 + L C_1 C_2}}.[/latex]
Let the voltage source be denoted by [latex]V_s[/latex]. At natural frequency, [latex]\omega_o[/latex], the voltage across the impedence [latex]Z_1[/latex] is,
[latex]V_1 = \frac{Z_1}{Z_1 + Z_2} \times V_s = j \frac{\frac{\omega_o L}{1 - \omega_o^2 L C_1}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \times V_s[/latex]
and the voltage across the impedence [latex]Z_2[/latex] is,
[latex]V_2 = \frac{Z_2}{Z_1 + Z_2} \times V_s = \left[ 1 - j \left( \frac{ \frac{\omega_o C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \right) \right] \times V_s = \left[ 1 - j \left( \frac{\omega_o C_2}{R \left[ \left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2 \right]^2} \right) \right] \times V_s[/latex]
Okay, now let [latex]V_s = 12, \ L = 0.1mH, \ C_1 = 100nF, \ C_2 = 80nF, \ R = 100 \Omega[/latex]. Using the formulae above I got,
[latex]\omega_o = 220479.275[/latex]Hz, [latex]V_1 = 12 \times (j 1.04364 \times 10^{3})[/latex], and [latex]V_2 = 12 \times (1 - j 1.04364 \times 10^{3})[/latex].
These mean [latex]|V_1| = 12.52368 \times 10^3[/latex] and [latex]|V_2| = 12.52368 \times 10^3[/latex]. But my simulation using SPICE, I got both [latex]|V_1|[/latex] and [latex]|V_2|[/latex] roughly around 12 volts only! Can someone tell me where I am wrong?
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