All categories

2 years ago

When does a circuit is said to be over loaded?

Physics

2 answers:

Vlad [161]2 years ago

8 0

Answer:

Every electric circuit in a wiring system must be protected against overloads. A circuit overload occurs when the amount of current flowing through the circuit exceeds the rating of the protective devices. The amount of current flowing in a circuit is determined by the load -- or the "demand" -- for current.

Explanation:

Hope this helps :)

Yakvenalex [24]2 years ago

8 0

Answer:

a circuit is said to be overloaded when there is a fluctuation in flow of current,when large amount of current which increases kinetic energy thus increasing heat energy thus incresing temperature so circuit melts which may prove fatal.

Explanation:

You might be interested in

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

VladimirAG [237]

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

<u>Explanation:</u>

Given data,

E= 3 ×10 ⁶ Δx=0.06/100

We have to find the minimum potential difference

E= -ΔV/Δx

ΔV=- E × Δx

ΔV =-3 ×10 ⁶ . 0.06/100

ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

6 0

3 years ago

Correct the following statement:

ankoles [38]

Answer:

It's the 3rd option

Explanation:

Wind is caused by the differences in air pressure on Earth's surface.

4 0

3 years ago

A charge of 5.67 x 10^-18 C is placed 3.5 x 10^-6 m away from another charge of -

Fudgin [204]

+ 1.58 e -15

Please hit thanks button! :)

5 0

1 year ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

Some machines do not multiply the force that is applied to them

Katyanochek1 [597]

Answer:

Machine Efficiency

Explanation:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work). The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent

5 0

2 years ago

When does a circuit is said to be over loaded?​

When does a circuit is said to be over loaded?