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The relation of peak current, peak voltage and capacitive reactance in alternating current is given by:

$$i_m=\frac{v_m}{X_c}$$

and $$X_c=\frac{1}{C\omega } \, .$$

So if we have a circuit with a capacitor and a bulb connected in series in an AC circuit, the bulb keeps glowing due the property of AC current. So if we reduce capacitance, $X_c$ increases.

So $i_m$ decreases (and $v_m$ stays constant), but if we rearrange the former equation can't we say that $v_m$ increases (and $i_m$ stays constant).

According to my text $i_m$ decreases (and $v_m$ stays constant). But how do we know that? Why can't it be the other way? Am I missing any concept?

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  • $\begingroup$ In English, only capitalize the first word of each sentence, and the names of people and places. $\endgroup$
    – DanielSank
    Mar 7, 2017 at 4:54

2 Answers 2

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I think that the problem here is that you haven't properly set up the circuit equations.

So if we have a circuit with a capacitor and a bulb connected in series in an AC circuit

Since the bulb and capacitor are series connected, the current through each is identical.

Denote the series current phasor as $I_s$. Assuming the source is an AC voltage source, it is true that the voltage across the source is constant.

But, by Kirchhoff's voltage law, we have

$$V_s = V_c + V_b$$

where the phasor voltages above are the source voltage, capacitor voltage and bulb voltage respectively.

By Ohm's law, we have

$$V_c = I_s \frac{1}{i\omega C}$$

and

$$V_b = I_s R_b $$

Thus,

$$I_s = \frac{V_s}{R_b + \frac{1}{i\omega C}}$$

Then, holding $V_s$ constant, see that decreasing $C$ decreases $I_s$.

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  • $\begingroup$ I'm not sure he is thinking about the bulb. The bulb seems to be in his text to demonstrate that an alternating current is flowing through the circuit. The confusion he seems to have is about the voltage or the current being constant. $\endgroup$ Oct 26, 2016 at 0:26
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So $i_m$ decreases (and $v_m$ stays constant), but if we rearrange the former equation can't we say that $v_m$ increases (and $i_m$ stays constant).

Generally when one is dealing with a circuit, the voltage supplied comes from somewhere such as a power supply. Power supplies usually supply a fixed voltage rather than a fixed current, which is why $v_m$ "stays constant". (There are also fixed-current power supplies in which the voltage alters, while the current stays constant. These are not common, though.)

According to my text $i_m$ decreases (and $v_m$ stays constant). But how do we know that? Why can't it be the other way? Am I missing any concept?

It certainly can be the other way around, but it depends on what else is connected to the circuit.

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