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In the laboratory you find an inductor of unknown inductance $L$ and unknown internal resistance $R$. Using a dc ohmmeter, an ac voltmeter of high impedance, a 1 microfarad capacitor, and a 1000 Hz signal generator, determine $L$ and $R$ as follows. According to the ohmmeter, $R$ is 35 ohms. You connect the capacitor in series with the inductor and the signal generator. The voltage cross both is 10.1 volts. The voltage across the capacitor alone is 15.5 volts. You not also, as a check, that the voltage across the inductor alone is 25.4 volts. How large is $L$? Is the check consistent?

Can someone tell me what's wrong with this logic (or if it's not wrong!)?

  • $V_B$ = voltage across both (10.1)

  • $\omega$ = angular frequency $(2\pi1000)$

  • C = 1 micro-farad ($1\ 10^{-6}$ F)

  • $V_C$ = 15.5

  • $V_L$ = 25.4

  • $R$ = 35

$$V_C/Z_C = V_B/(Z_R + Z_L - Z_C) = V_L/(Z_R + Z_L)$$

So:

$$15.5\omega C = 10.1/(R+\omega L-(1/(\omega L))) = 25.4/(R+\omega L)$$

Solving you get $L$ ~ .037 H.

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  • $\begingroup$ The impedance of a resistor is real while the impedance of an inductor or capacitor is imaginary. You're adding them together as if they are all real so the result cannot be correct. For example, your should write: $$|I| = \frac{|V_L|}{|Z_R + Z_L|} = \frac{|V_L|}{|R + j\omega L|} = \frac{|V_L|}{\sqrt{R^2 + (\omega L)^2}}$$ $\endgroup$ Apr 19, 2016 at 17:55
  • $\begingroup$ @AlfredCentauri Right that makes sense, thank you! I was forgetting that. So is it still correct for this problem to say $$|V_L|/|Z_R + Z_L| = |V_C|/|Z_C| = |V_B|/|Z_L + Z_R + Z_C|$$ $\endgroup$
    – Joe
    Apr 19, 2016 at 18:38
  • $\begingroup$ How can the voltage across the capacitor alone and across the inductor alone be bigger than the voltage across both? $\endgroup$ May 1, 2016 at 12:18
  • $\begingroup$ Because the voltage across the inductor is negative. $\endgroup$
    – Joe
    May 2, 2016 at 20:55

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