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An electric lamp having coil of negligible inductance connected in series with a capacitor and an AC source is glowing with certain brightness. How does the brightness of the lamp change on reducing the (i) capacitance, and (ii) the frequency ? What if the diaelectric slab is introduced between the capacitor.

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closed as too localized by Qmechanic, Manishearth, Emilio Pisanty, Sklivvz Dec 10 '12 at 10:16

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Hello Big Genius... It looks like some homework question..! –  Waffle's Crazy Peanut Sep 23 '12 at 14:21
    
Its not my homework, but i think question of this type come under homework questions. So , your welcome! –  BigSack Sep 24 '12 at 2:34
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1 Answer 1

up vote 1 down vote accepted

Reactance $ = X_c = \frac{1}{2\pi Cf}$

If you increase either the frequency or the capacitance the reactance will increase.

Impedance $= Z = R + iX$

Find the absolute value of impedance...

$|Z| = \sqrt{R^2 + i^2Z^2}$ therefore

$|Z| = \sqrt{ R^2 - Z^2}$

The impedance is a measure of the opposition of current. So decrease the impedance, increase the current.

$ZI = V$

So I believe that decreasing the capacitance or decreasing the frequency will increase the impedance and the bulb will be dimmer.

A dielectric produces a higher capacitance by allowing the capacitor to "hold" more charge. So it will increase the brightness by increasing the capacitance.

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Thanks! Enjoy your day. –  BigSack Sep 24 '12 at 2:36
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