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3 years ago

Kirchoff's Law states that, by the time current has returned to its source, all

Engineering

2 answers:

Alchen [17]3 years ago

6 0

Kirchoff's law states that by the time current has reached to its source all voltage must be used.

<u>Explanation:</u>

Kirchoff's voltage law states that the algebraic sum of all the voltages across a closed loop is zero. As the circuit is a closed conducting path, no energy is lost. ices All the voltage that is supplied from the voltage source is dropped at the devices in the circuit.

The potential difference maintained by the voltage source enables the flow of current across the circuit. As the supplied voltage drops at each elements of the circuit, all voltage is used by the time current has returned to the surface.

Alex17521 [72]3 years ago

4 0

Kirchoff's Law Kirchoff's Law states that, by the time current has returned to its source is explained in the following.

Explanation:

Kirchhoff's Current Law (KCL) is Kirchhoff's first law that deals with the conservation of charge entering and leaving a junction. ... In other words the algebraic sum of ALL the currents entering and leaving a junction must be equal to zero.

Kirchoff's laws apply for a given instant in time. So the voltages at a given moment around a loop will all sum to zero, or currents in a node sum to zero if you look at the instantaneous voltage and current. But they will be out of phase.

Kirchhoff Voltage Law states that ''The algebraic sum of all voltages (source voltage and voltage drops) is equal to zero around a close path''. This is called KVL ( Kirchhoff Voltage Law) equation. The source voltage is equal to the sum of all voltage drops.

Kirchhoff's Voltage Law (KVL) is Kirchhoff's second law that deals with the conservation of energy around a closed circuit path.

Kirchhoff's laws can be used to determine the values of unknown values like current, Voltage in the circuit. These laws can be applied on any circuit (with some limitation), and useful to find the unknown values in complex circuits and networks.

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A long, cylindrical, electrical heating element of diameter 10 mm, thermal conductivity 240 W/m·K, density 2700 kg/m3, and speci

Slav-nsk [51]

Answer:

Answer for the question is given in the attachment .

Explanation:

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3 years ago

Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

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4 0

2 years ago

Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo

rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

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3 years ago

On what frequency can you expect to monitor air traffic in and around<br> Lincoln Airport?

Phantasy [73]

Answer:

118.5

Explanation:

Hope this helps!

5 0

2 years ago

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

3 years ago