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1 year ago

Where are overcurrent protective devices normally installed in a branch circuit?

1 answer:

5 0

Overcurrent protective devices are normally installed in a branch circuit from where the conductors receive their supply.

<h3>What is resistance?</h3>

Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

V is the voltage,I is the current and R is the resistance

The vercurrent protective devices are normally installed in a branch circuit from where the conductors receive their supply.

Thus, In a branch circuit, overcurrent safety devices are often located from where the conductors get their supply.

Learn more about resistance from here, refer to the link;

brainly.com/question/14547003

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What velocity must a car with a mass of 1110 kg have in order to have the same momentum as a 2280 kg pickup truck traveling at 2

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49.3 m/s

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3 years ago

a person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18 m/s. The cliff is 52 m above the

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3 years ago

A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of

zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}$

$\dfrac{1}{f}=\dfrac{347}{5852}$

$f=16.86\ cm$

We need to calculate the focal length

Using formula of magnification

$m= \dfrac{-v}{u}$

Put the value into the formula

$2=\dfrac{v}{u}$

$v = -2u$

Using formula of for focal length

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}$

$\dfrac{1}{16.86}=\dfrac{1}{2u}$

$2u=16.86$

$u=\dfrac{16.86}{2}$

$u=8.43\ cm$

Hence, The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

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3 years ago