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ser-zykov [4K]
3 years ago
12

If temp. gets cold resistance in thermostat increases so voltage across it increases AND LED lights brighter. I understand every

thing except that for LED. It emits light when current passes through it, but if resistance increases how can current increase???​

Physics
1 answer:
Valentin [98]3 years ago
5 0

Normally, when something gets colder, its electrical resistance gets smaller.  This is true of component-A in the drawing ... a simple resistor.

The component labeled 'B' has a strange and unusual symbol, and it's not a simple resistor.  It's a "thermistor".  The word "thermal" always has something to do with heat, and "thermistor" comes from "thermal resistor.  These things can be manufactured either way ... using different materials, a thermistor can be manufactured so that its resistance goes UP, or goes DOWN, or doesn'tchange when it gets colder.  I'm pretty sure that's what's going on here.

When this circuit gets colder, resistance-A gets smaller, but resistance-B either gets bigger OR doesn't change.  Either way, the voltage across B increases.  Since the LED is connected directly across B, the current through it depends on that voltage, so the LED gets more current, and becomes brighter, when A and B both get colder.

This circuit could actually be a very useful device.  If you took out the LED and put a voltmeter in its place, then the reading on the voltmeter would tell you the temperature of wherever you put the two components A and B.

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3 years ago
1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
labwork [276]

Answer:

1.35 kJ  

Explanation:

KE = ½mv² = ½ × 0.030  kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

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What does an atomic number represent in an atom?
Bezzdna [24]

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3 years ago
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Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its
kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2},

Why does this formula work?

By Faraday's Law of Induction, the EMF \epsilon induced in a coil (one loop) is equal to the rate of change in the magnetic flux \Phi through the coil.

\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t)).

Finding the average EMF in the coil is similar to finding the average velocity.

\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt.

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0).

Hence the equation

\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}.

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in \Phi(t) won't matter.

Apply this formula to this question. Note that \Phi, the magnetic flux through the coil, can be calculated with the equation

\Phi = B \cdot A \cdot N \; \sin{\theta}.

For this question,

  • B = \rm 16\; mT = 16\times 10^{-3}\; T is the strength of the magnetic field.
  • A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2} is the area of the coil.
  • N = 1 is the number of loops in the coil.
  • \theta is the angle between the field lines and the coil.
  • At \rm 0\;s, the field lines are parallel to the coil, \theta = 0^{\circ}.
  • At \rm 0.7\; s, the field lines are perpendicular to the coil, \displaystyle \theta = 90^{\circ}.

Initial flux: \Phi(0)= 0.

Final flux: \Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb.

Average EMF, which is the same as the average rate of change in flux:

\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V.

8 0
3 years ago
Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne
VLD [36.1K]

Answer:

The the maximum emf is 2.52\times10^{-11}\ V

Explanation:

Given that,

Magnetic field B = 1.40\times10^{-3}\ T

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

\epsilon=NBA\omega

Where, N = number of turns

B= magnetic field

A = area

Put the value in to the formula

\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60

\epsilon=2.52\times10^{-11}\ V

Hence, The the maximum emf is 2.52\times10^{-11}\ V

5 0
3 years ago
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