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mr_js
mr_js
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One option would be to form a series RL circuit and apply a sinusoidal voltage, vi\$v_i\$, of a given frequency, \$f\$. Then measure the phase difference between the input and output voltage. From the voltage divider equation vo/vi = jwL/(R+jwL)\$\frac{v_{o}}{v_{i}}=\frac{j\omega L}{R+j\omega L}\$ the phase difference is equal to theta = 90 - atan(wL/R)\$90^{\circ}-\arctan(\frac{\omega L}{R})\$. Thus you can solve for L. You

You can make this even simpler by varying the resistance and/or frequency of the sinusoidal source until the phase shift between input and output voltages is exactly 45 deg\$45^{\circ}\$. At this point, the reactance of the inductor is equal to the resistance. Therefore L = R/(2.pi and the inductance is given by \$L = R/(2\pi f)\$.f)

One option would be to form a series RL circuit and apply a sinusoidal voltage, vi, of a given frequency. Then measure the phase difference between the input and output voltage. From the voltage divider equation vo/vi = jwL/(R+jwL) the phase difference is equal to theta = 90 - atan(wL/R). Thus you can solve for L. You can make this even simpler by varying the resistance and/or frequency of the sinusoidal source until the phase shift is exactly 45 deg. At this point, the reactance of the inductor is equal to the resistance. Therefore L = R/(2.pi.f)

One option would be to form a series RL circuit and apply a sinusoidal voltage, \$v_i\$, of a given frequency, \$f\$. Then measure the phase difference between the input and output voltage. From the voltage divider equation \$\frac{v_{o}}{v_{i}}=\frac{j\omega L}{R+j\omega L}\$ the phase difference is equal to \$90^{\circ}-\arctan(\frac{\omega L}{R})\$. Thus you can solve for L.

You can make this even simpler by varying the resistance and/or frequency of the sinusoidal source until the phase shift between input and output voltages is exactly \$45^{\circ}\$. At this point, the reactance of the inductor is equal to the resistance and the inductance is given by \$L = R/(2\pi f)\$.

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mr_js
mr_js
  • 1.1k
  • 6
  • 13

One option would be to form a series RL circuit and apply a sinusoidal voltage, vi, of a given frequency. Then measure the phase difference between the input and output voltage. From the voltage divider equation vo/vi = jwL/(R+jwL) the phase difference is equal to theta = 90 - atan(wL/R). Thus you can solve for L. You can make this even simpler by varying the resistance and/or frequency of the sinusoidal source until the phase shift is exactly 45 deg. At this point, the reactance of the inductor is equal to the resistance. Therefore L = R/(2.pi.f)

One option would be to form a series RL circuit and apply a sinusoidal voltage, vi, of a given frequency. Then measure the phase difference between the input and output voltage. From the voltage divider equation vo/vi = jwL/(R+jwL) the phase difference is equal to theta = 90 - atan(wL/R). Thus you can solve for L.

One option would be to form a series RL circuit and apply a sinusoidal voltage, vi, of a given frequency. Then measure the phase difference between the input and output voltage. From the voltage divider equation vo/vi = jwL/(R+jwL) the phase difference is equal to theta = 90 - atan(wL/R). Thus you can solve for L. You can make this even simpler by varying the resistance and/or frequency of the sinusoidal source until the phase shift is exactly 45 deg. At this point, the reactance of the inductor is equal to the resistance. Therefore L = R/(2.pi.f)

Source Link
mr_js
mr_js
  • 1.1k
  • 6
  • 13

One option would be to form a series RL circuit and apply a sinusoidal voltage, vi, of a given frequency. Then measure the phase difference between the input and output voltage. From the voltage divider equation vo/vi = jwL/(R+jwL) the phase difference is equal to theta = 90 - atan(wL/R). Thus you can solve for L.