All categories

3 years ago

What is an electric circuit?

Physics

2 answers:

klemol [59]3 years ago

7 0

<h3>Answer: any path that allows electrons to flow</h3>

An electrical circuit is a path in which electrons from a voltage or current source flow. ... The part of an electrical circuit that is between the electrons' starting point and the point where they return to the source is called an electrical circuit's "load".

attashe74 [19]3 years ago

3 0

A type of circuit where electricity can flow through more than one path. voltage. the force or push by which electricity is pushed through a circuit. electric. having to do with, making, or using electricity.

You might be interested in

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

3 years ago

You are given three resistors with the following resistances: R1 = 6.32 Ω, R2 = 8.13 Ω, and R3 = 2.29 Ω. What is the largest equ

andreyandreev [35.5K]

Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

Explanation:

Adding Resistances in serie is the way to build de largest equivalent value possible.

Rt= R1+R2+R3

Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

5 0

3 years ago

An object is pulled northward with a force of 10 n and southward with a force of 15 n. the magnitude of the net force on the obj

Anastaziya [24]

The correct option is (b) 5n

As a result, there is a net downward force of 5N operating on the object.

The resultant force is the force that results from adding the vector sums of all the forces operating on an item. The combined action of all the acting forces on the object produces the same effect as the resulting force. When determining the resulting force, the direction of the forces must be taken into account.

Given;

The northward force is Fn = 10N

The southward force is Fs = 15N

Required;

The net force on the mobile phone is Fnet = ?N

The object's weight exerts downward pressure, and upward resistance exerts upward pressure. The vector sum of these two forces will be the net force.

Fnet = Fs - Fn (Considering the direction downward as positive)

Fnet= 15N - 10N

Fnet = 5N

As a result, there is a net downward force of 5 N operating on the object.

Learn more about the Force with the help of the given link:

brainly.com/question/7362815

#SPJ4

6 0

1 year ago

A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no

padilas [110]

Explanation:

The weight of the car is equal to, $W_c=m\times g$ ...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, $F_N=mg\ cos\theta$

$F_N=mg\ cos(13)$ .............(2)

The horizontal component of the force is, $F_H=mg\ sin\theta$

Taking ratio of equation (1) and (2) as :

$\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}$

$\dfrac{F_N}{W_c}=cos(13)$

$\dfrac{F_N}{W_c}=0.97$

$\dfrac{F_N}{W_c}=\dfrac{97}{100}$

Hence, this is the required solution.

5 0

3 years ago

] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0

3 years ago