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Alex Ar [27]
3 years ago
6

A 10-μF capacitor in an LC circuit made entirely of superconducting materials ( R = 0 Ω ) is charged to 100 μC. Then a supercond

ucting switch is closed. At t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. At a later time, Vab = +10V . At that time, Vdc is
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

Vdc=10V

Explanation:

in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,

if the loop remains closed for  a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

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What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point
VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

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2 years ago
A conducting rod moves perpendicularly through a uniform magnetic field. How does the emf change if the magnetic field is increa
Marysya12 [62]

Answer:

c) It increases by a factor of 8

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V = -d(BA)/dt  or -BA/t

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The EMF has increased by a factor of 8.

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