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3 years ago

Shown below is a parallel circuit of two light bulbs.

1 answer:

3 0

The characteristics of parallel circuits allow to find the correct answer for the question about what happens when a light bulb is:

D) Bulb 2 will stay lit and the light source will be stronger.

Electrical circuits have two forms and connection:

• Serie. In this case there is only one path through which current can circulate the current.

• Parallel. There are several roads where the current can circulate.

In a parallel circuit, the total current in the circuit is the sum of the current in each branch, the voltage in all branches remains constant.

Let's analyze the claims:

A) False. The bulbs are independent, what happens to one does not affect the other.

B) False. Both bulbs were on, so the second bulb should not be affected.

C) False. When burn a light bulb, there is more current in the other branch, The opposite happens.

D) True. When burn a light bulb there is more current in the other branch and therefore the light bulb 2 can dissipate more power by shining a little brighter.

In conclusion, using the characteristics of parallel circuits we can find that the correct answer for the question about what happens when a light bulb burns out is:

D) Bulb 2 will stay lit and the light source will be stronger.

Learn more about parallel circuits here: brainly.com/question/23088951

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Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

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n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

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3 years ago

Which type of radiation has the highest penetrating power?

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electromagnetic.........

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3 years ago

Solve the problem.

gulaghasi [49]

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