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2 years ago

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A device is turned on and 3.20 A flows through it 0.130 ms later. What is the self-inductance of the device (in mH) if an induce

d 140 V emf opposes this

1 answer:

jeka57 [31]2 years ago

8 0

Answer:

the self-inductance of the device is 5.69 mH

Explanation:

Given;

change in current, ΔI = 3.2 A

change in time, Δt = 0.13 ms = 0.13 x 10⁻³ s

induced emf, E = 140

The self-inductance is calculated as;

$E = L\frac{\Delta I}{\Delta t} \\\\where;\\\\L \ is \ the \ self-inductance\\\\\L = \frac{E\Delta t}{\Delta I} \\\\L =\frac{140 \times 0.13 \times 10^{-3}}{3.2} \\\\L = 0.00569 \ H\\\\L = 5.69 \ mH$

Therefore, the self-inductance of the device is 5.69 mH

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Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

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third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

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is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

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For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

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a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

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$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

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$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

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