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2 years ago

If a 75 W lightbulb operates at a voltage of 120 V, what is the current in the

Physics

1 answer:

zheka24 [161]2 years ago

4 0

Answer:

Option D. 0.63 A

Explanation:

The following were obtained from the question:

P = 75W

V = 120 V

I =?

But,

P = IV

I = P / V

I = 75 / 120

I = 0.63A

The current in the bulb is 0.63A.

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The ideal gas constant is a proportionality constant that is added to the ideal gas law to account for pressure (P), volume (V), moles of gas (n), and temperature (T) (R). R, the global gas constant, is 8.314 J/K-1 mol-1.

According to the Ideal Gas Law, a gas's pressure, volume, and temperature may all be compared based on its density or mole value.

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1 month ago

If a truck travels 100 miles in 2 hours, what is its speed?

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Write a short paragraph that explains the central idea of the article. Use at least two details from the article to support your

zhannawk [14.2K]

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1 year ago

A baseball player slides into second base. What happens to his momentum?

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Due to friction from sliding against the ground, the player decelerates in 1 direction. Thus his momentum decreases.

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With two identical light bulbs and two identical batteries how would you arrange in order to get the maximum total power to the

Norma-Jean [14]

Answer:

The batteries would be connected in series while the bulbs would be connected in parallel

Explanation:

Power (W) = VI

where V = voltage, I = current and R = resistance

from V = IR , I = V/R

Power (W) now becomes = V (V/R) = $\frac{V^{2} }{R}$

Power (W) = $\frac{V^{2} }{R}$

from the above equation, power is directly dependent on voltage, hence the voltage has to be high for the power to be high and the power is also inversely dependent on the resistance (in this case the bulbs which act as the load)

We have to batteries, when batteries are connected in series the total voltage becomes the summation of the two voltages hence giving a higher voltage and when they are connected in parallel their voltage remains the same. Since we want to get higher voltage we will connect the two batteries in series.
we have two bulbs which are the resistance here, from the equation above the power is inversely dependent on the resistance so we would need its value to be minimal. When resistance is connected in series the resistance individual will be added to get the total resistance, hence the total resistance will be high but when the resistors are arranged in parallel you get the total resistance by applying the formula $\frac{R1R2}{R1+R2}$ which will give us a lower resistance. Hence we would connect the bulbs in parallel.

Take note that the power from this connection should not exceed the bulbs power rating so as to avoid damage of the bulbs.

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2 years ago