Question

A neon lamp is connected to a voltage a.c source. The voltage is gradually increased from zero volt. It is observed that the neon flash is at 50 V. the a.c. source is now replaced by a variable dc source and the experiment is repeated. The neon bulb will flash at:
A) 50V
B) 70V
C) 100V
D) 35V

Answer 1

Hint:The RMS value tells us the equivalent value for dc source. RMS stands for root mean square this RMS value helps us to find the equivalent value of the dc source if in a circuit a.c. source is replaced by a dc source. In the dc source there is no presence of phase.

Formula used:The formula of the RMS voltage is given by ${V_{dc}} = {V_{a.c.}} \times \sqrt 2 $ where $V$ is the voltage of the dc source and ${V_o}$ is the voltage of the a.c. source.

Complete step by step answer:
It is given that a lamp is initially connected by an a.c. source and at 50V the lamp used to glow but after the replacement of a.c. source into the dc source we get that the voltage at which the lamp glows will change. The RMS i.e. root mean square value tells us how the conversion will take place and the reason for that is there is a sinusoidal wave in the a.c. source and after replacement of the voltage source from a.c. to dc source there is no sinusoidal variation in the voltage source.
As the voltage of the dc source is given by,
${V_{dc}} = {V_{a.c.}} \times \sqrt 2 $
Where ${V_{dc}}$ is the voltage of the dc source and ${V_{a.c.}}$ is the voltage of the a.c. source.
${V_{dc}} = {V_{a.c.}} \times \sqrt 2 $
$ \Rightarrow {V_{dc}} = 50 \times \sqrt 2 $
$ \Rightarrow {V_{dc}} = 50\sqrt 2 $
$ \Rightarrow {V_{dc}} = 50 \times 1 \cdot 414$
$ \Rightarrow {V_{dc}} = 70.7V$
$ \Rightarrow {V_{dc}} \approx 70.V$

The correct answer for this problem is option B.

Note: The change from a.c. source to dc source takes place then the voltage has some sinusoidal waves but as the dc source is applied then the voltage becomes a constant supply and no increase and decrease of the voltage with respect to the phase takes place.