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?
Best Answer
I think that the problem here is that you haven't properly set up the circuit equations.
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 $I_s$.