Question

A 20 volts 5 watt lamp is used in a.c. main of 220 volts 50 c.p.s. calculate the
(i) capacitance of capacitor.
(ii) inductance of inductor, to be put in series to run the lamp.
(iii) what pure resistance should be included in place of the above device so that the lamp can run on its voltage?
(iv) which of the above arrangements will be more economical and why?

Solution

Given,
Voltage of lamp, V = 20 V
Power of lamp = 5 W
Rms voltage, V_rms = 220 V
Hence,
Current required by the lamp,

I =

\frac{wattage}{voltage} = \frac{5}{20} = 0.25 amp .

Resistance of the lamp

R = \frac{voltage}{current} = \frac{20}{0.25} = 80 ohm

so for proper running of the lamp, the current through the lamp should be 0.25 amp.

(i) When the condenser C is placed in series with lamp, then

Z = \sqrt{[R^{2} + {(\frac{1}{ωC})}^{2}]}

The current through the circuit
I =

\frac{200}{\sqrt{[R^{2} + {(\frac{1}{ωC})}^{2}]}} = 0.25

\frac{200}{\sqrt{(80)^{2} + (\frac{1}{4 π^{2} \times 50^{2} + C^{2}})}} = 0.25

Solving it for C, we get

C = 4.0 \times 10^{- 6} F = 4.0 μF

(ii) When inductor L henry is placed in series with the lamp, then

Z = \sqrt{[R^{2} + (ωL)^{2}]}

\frac{200}{\sqrt{[R^{2} + (ωL)^{2}]}} = 0.25

\frac{200}{\sqrt{[(80)^{2} + (4 π^{2} \times 50^{2} \times L^{2})]}} = 0.25

Solving it for L, we get L = 2.53 henry.

(iii) When resistance R ohm is placed in series with lamp of resistance R, then

\frac{200}{R + r} = 0.25 \frac{200}{80 + r} = 0.25

\Rightarrow

r = 720 ohms

(iv) It will be more economical to use inductance or capacitance in series with the lamp to run it as it consumes no power while there would be dissipation of power when, resistance is inserted in series with the lamp.

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