a lamp is connected in series with a capacitor .predict your observations for 1)dc and 2)ac connection. What happens in each case if dielectric slab is introduced in the capacitor?
Answers
For AC connection, Lamp will shine as the current is finite i.e.
Now, if the capacitance is reduced, the (1/ωC) factor in the denominator increases and hence the current decreases. Thus brightness of the lamp reduces.
Hey !!
When a dc source is connected to a capacitor, the capacitor gets charged and after charging, no current flows in the circuit and the lamp will not glow. There will no change even if C is reduced. With an ac source, the capacitor offers capacitive reactance and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase the reactance and the lamp will shine less brightly than before.
DETAILED EXPLANATION
In case of dc, the frequency is zero, so the capacitive reactance Xc = 1/2πfc is infinite, hence there will no flow of current in the circuit and the lamp will not glow. If the capacitance of the capacitor is reduced, there will be no change in the circuit which makes the lamp to remain in similar position.
In case of ac, the capacitors give finite reactance which allows the current to flow making the lamp to glow. When capacitance C of the capacitor gets reduced, there will be an increase in the reactance Xc = 1/2πfc which leads to decrease in circuit current making the lamp to glow dim.