Question

(a) The bulb is in series with a capacitor along with the a.c. mains. The bulb lights with some intensity. What happens to the intensity of the bulb (i) by doubling the capacitance (ii) by doubling the frequency of a.c. (iii) if the AC supply is replaced with DC? Give reasons in support of your answer. (3marks) (b) A 2032 resistor is in series with a 30uF capacitor along with 24 V a.c.power supply. What will be the frequency of the power supply if the circuit current is 3 mA. (3marks) (c) Draw the graphs between the following physical quantities (1mark) (i) current through resistor with frequency (ii) voltage and current through resistor

Answer 1

Capacitive reactance is given by

Хc jwC

Therefore, the total impedence of the series RC circuit is

$Z=\sqrt{R^{2}+X^{2}_{C}}=\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}$

The intensity of the bulb (B) is proportional to the heat generated in the bulb. If the voltage of the AC source is V, then you can write

2 V BX1²R (1) XR Z

R ..B R2 + 1

(a) (i) Doubling the capacitance means intensity will increase according to above equation.

(ii) Doubling the frequency means intensity will increase according to above equation.

(iii) When the voltage reaches a threshold value, a current flows through the bulb that dramatically reduces its resistance, and the capacitor (capacitance C and time constant $\tau$ ) discharges through the bulb as if the battery and charging resistor were not there. Once discharged, the process will start again, with a flash period of $\tau$ .

(b) The circuit current (I) can be expressed as for AC RC circuit

$I=\frac{V}{Z}=\frac{V}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}$

Here, in your problem, R = 20 ohm, C = 30 $\mu F$ = 30 x 10-6 F, I = 3 mA = 3 x 10-3 A = 0.003 A, V = 24 Volt. Therefore,

$0.003=\frac{24}{\sqrt{20^{2}+\frac{1}{\omega^{2} (30\times 10^{-6})^{2}}}}$

$\therefore 20^{2}+\frac{1}{\omega^{2} (30\times 10^{-6})^{2}}=64\times 10^{6}$

$\therefore \omega = 4.16\ rad/s$

$\therefore f = 0.663\ Hz$

(c) (i) The current vs frequency will look like (Frequency along X-axes and Current along Y-axes)

. HER

(ii) THe current and voltage variation of resistor will be same as the figure above with frequency, but the data value indicated in the Y-axes will change accordingly.