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Delicious77 [7]
3 years ago
9

1) A capacitor consists of two metal surfaces separated by an electrical insulator with no electrically conductive path through

it. Why does a current flow in a resistor-capacitor circuit when the switch is closed?
Physics
1 answer:
ki77a [65]3 years ago
7 0

Answer:

Current flows in a resistor-capacitor circuit because of the varying electric field across the plates of a capacitor induced by an AC voltage source <em>(displacement current)</em>

Explanation:

In a capacitor, current does not flow the same way it does in a circuit, that is through conduction. This is because there is a highly resistive material in between the plates of the capacitor. Rather current flows through a phenomenon called displacement current.

Because of change in charge accumulation with time above the plates, the electric field changes causing the displacement current.

Displacement current arises due to the flow of electrons as a result of the varying magnetic fields set up on the plates of the capacitor when supplied with an AC voltage. It is important to note that a DC voltage does not induce any displacement current.

<em>Through this, phenomenon discovered by Maxwell,  current is able to flow in a resistor-capacitor circuit despite the absence of an electrically conductive path through the plates.</em>

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><span>It can travel through vacuum. 

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Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

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we evaluate  V₀ = 4 V

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