What happens when the supply voltage to a light bulb is increased?
A. The bulb glows brighter
B. The bulb consumes less power
C. The bulb consumes more power
D. The power consumption does not change.
Which of the above statements are true?
Answer
608.1k+ views
Hint: As we know that voltage is directly proportional to the current flowing. If the voltage increases then the brightness will also increase and if the voltage decreases then the brightness will also decrease.
Formula Used:
$P=I \times V$ ------- (1)
Also we know that power is also\[ = {I^2}R\]
Here $P$ is the power and $I$ is the current flowing and $V$ is the voltage and $R$ is the resistance.
Complete step by step answer:
According to the given question, Voltage of the supply is increasing and we know that Power\[ = I \times V\], Here $I$ is current and $V$ is voltage.
$P=I \times V$ ------- (1)
Also we know that power is also\[ = {I^2}R\] ----- (2)
If we compare equation (1) and (2), We get,
\[I \times V = \dfrac{{{I^2}}}{R}\]
From this equation, we get-
\[I = \dfrac{V}{R}\]
As, from this equation it is clear that the current is directly proportional to the voltage supply. If current is more than the bulb will glow brighter and more power will be consumed.
So, option A and C are correct for this given statement.
Note:As we know, power is equal to the rate at which energy is delivered. SI unit of power is \[watt\] which is equal to\[J/\sec \]. If we want to write power formula, it will be-
\[P = VI = {I^2}R = \dfrac{{{V^2}}}{R}\]
Here $P$ is power consumed, $V$ is voltage, $R$ is the resistance and $I$ is the current flowing in the circuit. It is clearly visible that power is depending on the voltage and current directly. If voltage and current is increasing, then the power increases.
Formula Used:
$P=I \times V$ ------- (1)
Also we know that power is also\[ = {I^2}R\]
Here $P$ is the power and $I$ is the current flowing and $V$ is the voltage and $R$ is the resistance.
Complete step by step answer:
According to the given question, Voltage of the supply is increasing and we know that Power\[ = I \times V\], Here $I$ is current and $V$ is voltage.
$P=I \times V$ ------- (1)
Also we know that power is also\[ = {I^2}R\] ----- (2)
If we compare equation (1) and (2), We get,
\[I \times V = \dfrac{{{I^2}}}{R}\]
From this equation, we get-
\[I = \dfrac{V}{R}\]
As, from this equation it is clear that the current is directly proportional to the voltage supply. If current is more than the bulb will glow brighter and more power will be consumed.
So, option A and C are correct for this given statement.
Note:As we know, power is equal to the rate at which energy is delivered. SI unit of power is \[watt\] which is equal to\[J/\sec \]. If we want to write power formula, it will be-
\[P = VI = {I^2}R = \dfrac{{{V^2}}}{R}\]
Here $P$ is power consumed, $V$ is voltage, $R$ is the resistance and $I$ is the current flowing in the circuit. It is clearly visible that power is depending on the voltage and current directly. If voltage and current is increasing, then the power increases.
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