
The neon lamp is connected to a voltage a.c source. The voltage is gradually increased from zero volt. It is observed that neon flashes at $50V$. The a.c source is now replaced by the variable dc source and experiment is repeated. What will be the voltage at which the neon bulb will flash?
A) $50V$
B) $70V$
C) $100V$
D) $35V$
Answer
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Hint: The RMS stands for root mean square and helps us to find the equivalent value of dc source if in the circuit a.c source is replaced by the dc source. When the source is dc then there is no presence of phase.
Formula used:
In this question we have to calculate the value of RMS voltage which is given by –
${V_{dc}} = {V_{a.c}} \times \sqrt 2 $
where, ${V_{dc}}$ is the voltage of a dc source and ${V_{a.c}}$ is the voltage of a source.
Complete step by step solution:
According to the question, it is given that initially the lamp is connected to the a.c source. The lamp glows at the voltage of $50V$ and when the a.c source is replaced with dc source the voltage at which the lamp glows will change. So, we need to find the RMS value of the voltage. The RMS value tells us that how the conversion will take place due to sinusoidal wave in the a.c source and when the voltage source is replaced from a.c to dc source there is no sinusoidal variation in the voltage source.
The voltage of the dc source is given by –
${V_{dc}} = {V_{a.c}} \times \sqrt 2 $
where, ${V_{dc}}$ is the voltage of a dc source and ${V_{a.c}}$ is the voltage of a source.
Putting the value of voltage at a.c source –
$
\Rightarrow {V_{dc}} = 50 \times \sqrt 2 \\
\Rightarrow {V_{dc}} = 50\sqrt 2 \\
\Rightarrow {V_{dc}} = 50 \times 1.414 \\
\Rightarrow {V_{dc}} = 70.7V \\
\Rightarrow {V_{dc}} \approx 70V \\
$
Hence, the voltage at dc source is $70V$.
Therefore, the correct option is (B).
Note: When the a.c source is replaced by the dc source then voltage has sinusoidal waves but when dc source is applied then voltage becomes constant and there is no increase or decrease of voltage with respect to the phase taking place.
Formula used:
In this question we have to calculate the value of RMS voltage which is given by –
${V_{dc}} = {V_{a.c}} \times \sqrt 2 $
where, ${V_{dc}}$ is the voltage of a dc source and ${V_{a.c}}$ is the voltage of a source.
Complete step by step solution:
According to the question, it is given that initially the lamp is connected to the a.c source. The lamp glows at the voltage of $50V$ and when the a.c source is replaced with dc source the voltage at which the lamp glows will change. So, we need to find the RMS value of the voltage. The RMS value tells us that how the conversion will take place due to sinusoidal wave in the a.c source and when the voltage source is replaced from a.c to dc source there is no sinusoidal variation in the voltage source.
The voltage of the dc source is given by –
${V_{dc}} = {V_{a.c}} \times \sqrt 2 $
where, ${V_{dc}}$ is the voltage of a dc source and ${V_{a.c}}$ is the voltage of a source.
Putting the value of voltage at a.c source –
$
\Rightarrow {V_{dc}} = 50 \times \sqrt 2 \\
\Rightarrow {V_{dc}} = 50\sqrt 2 \\
\Rightarrow {V_{dc}} = 50 \times 1.414 \\
\Rightarrow {V_{dc}} = 70.7V \\
\Rightarrow {V_{dc}} \approx 70V \\
$
Hence, the voltage at dc source is $70V$.
Therefore, the correct option is (B).
Note: When the a.c source is replaced by the dc source then voltage has sinusoidal waves but when dc source is applied then voltage becomes constant and there is no increase or decrease of voltage with respect to the phase taking place.
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