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In the figure below, enter image description here The current intensity is 4A, and V1 reads 80V, and V2 reads 64V. What is the reading of V3? In another way what is the equivalent voltage for an inductor and a capacitor? In my textbook, it says that V3 reads 16V, but I don't understand why the voltage across Xc and Xl together would be:

V1-V2=V3= 80-64=16.

Could you please explain it for me?

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4A flows through the resistor, inductor and capacitor and its phase is common to all three.

The voltage drop across the resistor = 4*R and is in phase with the current.

The voltage drop across the inductor leads the current by 90 degrees and will have the value (I * XL) = 4 * 20 = 80V

The voltage across the capacitor lags the current by 90 degrees and will have the value (I * Xc) = 4 * 16 = 64V

The voltage across the inductor and the capacitor are 180 degrees apart or antiphase so they cancel each other out leaving a nett voltage of 80 - 64 = 16V (the voltage measured by the meter).

This voltage is mainly due to the inductor so it will lead the current by 90 degrees.

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For the inductance your calculation is correct, as you have $$Z_L = jX_L$$ However, for the capacitance you should remember the sign of the impedance: $$Z_C = -jX_C$$ Both capacitors and inductors are frequency dependent, so if you would just put a DC voltage source over the circuit you should expect different voltages. The inductor will look like a short with 0Ω impedance and the capacitor will be an infinitely high impedance.

The reactance and inductance values that are given are for a certain frequency and moment in time.

The sign in the reactance formula originates from the frequency domain formula for the impedance of a capacitor: $$Z_C = \frac{1}{j\omega C}$$ Rewriting this in the reactive part would result in $$Z_C = -j \frac{1}{\omega C}$$ where $$\frac{1}{\omega C} = X_C$$

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  • \$\begingroup\$ Does "Z" in your answer refer to the voltage? Becuase I use it to refer to the impedance. \$\endgroup\$ – Asmaa Feb 5 '17 at 19:18
  • \$\begingroup\$ Z is the electrical impedance, X is called the reactance. So actually the impedance is negative. \$\endgroup\$ – Douwe66 Feb 5 '17 at 19:22
  • \$\begingroup\$ OK, but I thought the impedance is the equivalent resistance of more than one component (e.g. an inductor and a resistance). So, I think the resistance of the capacitor will be represented by the reactance Xc. Could you please give me more explanation about the negative sign you used in the formula? \$\endgroup\$ – Asmaa Feb 5 '17 at 19:27
  • \$\begingroup\$ @Douwe66 It is wrong to say the impedance is negative. The reactance and impedance of the inductor and capacitor are complex (imaginary, to be specific) so 'positive' and 'negative' do not apply (they are off the number line, so to speak or not in \$\mathbb{R}\$). \$\endgroup\$ – Spehro Pefhany Feb 5 '17 at 19:39
  • \$\begingroup\$ I added a bit more in my answer. The impedance is indeed the equivalent resistance, but it is frequency dependent. It is by convention that the sign is put with the capacitor. \$\endgroup\$ – Douwe66 Feb 5 '17 at 19:43
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enter image description here

Arthough arbitrary we consider L = +ve reactance and C =-ve reactance where voltage lags current. The negative sign is a rule of complex impedance for sine cosine phase shifts in electronics +/-90 deg V/I phase shifts only in pure reactive parts. such that series LC are always 180 deg apart but their impedance changes with f.

Thus the voltages cancel.

The rules of j ( or i symbol in pure math which was changed to prevent confusion with current) were discovered by Pyramid design Engineers and not discovered again until Napoleonic days by Descartes j=√ -1 and 1/j=-j fits the way LC impedances affect voltage with amplitude cancellation using a shared current. When the impedances are exactly equal, the voltage cross LC becomes zero (0) and a sweep frequency would show a notch filter response past this "natural" resonant frequency.

for fun read. enter image description here

Later you will find this is a series resonant circuit with peak current = V/R at resonant frequency and zero voltage across LC. a.k.a. Voltage notch filter or (current peak resonant circuit)

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