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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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You may make the following measurements of an object 42kg and 22m3. What would the objects density be?

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Density=mass/volume
to find the density
mass=42kg
volume=22m3
so density=42/22
density=1.9Kgm3

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3 years ago

Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older? A) Fossil A B) Fossil B C)

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Fossil B is older the lower they found the fossil the older it is

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2 years ago

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An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

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$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

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3 years ago

Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C

Studentka2010 [4]

Answer:

did u get it?

Explanation:

frfr

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3 years ago

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Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

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Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$