Question

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? justify your answer.

Answer 1

Hey mate ^_^

#Be Brainly❤️

Answer 2

Hello mate here is your answer.

Reactance =Xc=12πCf=Xc=12πCf

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

Impedance =Z=R+iX=Z=R+iX

Find the absolute value of impedance...

|Z|=R2+i2Z2−−−−−−−−√|Z|=R2+i2Z2 therefore

|Z|=R2−Z2−−−−−−−√|Z|=R2−Z2

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

ZI=VZI=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.

Hope it helps you