Question

A photocell connected in an electrical circuit is placed at a distance 'd' from a source of light. As a result, current I flows in the circuit. What will be the current in the circuit when the distance is reduced to d/3?

a) I
b) 6I
c) 9I
d) 1/3 I

Answer 1

Answer:1/4 Explanation:Intensity will be reduced as inversely proportion to distance. ... A photo-cell connected in an electrical circuit is placed at distance d from a source of light as a result current I flow in the circuit what will be the ... of 3 m. if a bus is located at 5m from this mirror find the position natu.

Answer 2

c) 9I

Explanation:

As the current of a photoelectric circuit depends upon the intensity of the light wave and as current is directly proportional to the intensity of light.

This intensity of light is inversely proportional to the square of the distance between the area on which it is projected and light source.

So,

$I\propto i$

where:

i = intensity of light

I = current

&

$i\propto \frac{1}{d^2}$

where:

$d=$ distance between the plane and the light source

New photo-current

$\therefore I'\propto \frac{1}{(\frac{d}{3})^2 }$

$I'=\frac{9}{d^2}$

$I'=9I$

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TOPIC: photoelectric current

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