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3 years ago

What is the difference between a battery losing charge (discharging) and a capacitor losing charge (discharging)?

1 answer:

6 0

The capacitor takes the charge from the battery but the electromotive force EMF of the battery remains same. But, during discharging, the internal resistance of the battery decreases because the charge moves out of the battery to the capacitor.
But, when the capacitor discharges its emf, decreases with time.

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The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r

pogonyaev

Given:

L = 1 mH = $1\times 10^{-3}$ H

total Resistance, R = 11 $\Omega$

current at t = 0 s,

$I_{o}$ = 2.8 A

Formula used:

$I = I_{o}\times e^-{\frac{R}{L}t}$

Solution:

Using the given formula:

current after t = 0.5 ms = $0.5\times 10^{-3} s$

for the inductive circuit:

$I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}$

$I = 2.8\times e^-5.5$

I =0.011 A

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3 years ago

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andrey2020 [161]

Answer: I would say it would be 3.9

but i believe there is something missing shorty.

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3 years ago

How do land and sea breeze work?

Studentka2010 [4]

by the wind and air flow in the wind

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3 years ago

What is meaning of phase

algol [13]

Answer:

phares

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2 years ago

Read 2 more answers

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago