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Gekata [30.6K]
3 years ago
11

A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre

nt in the capacitor at t
Physics
1 answer:
disa [49]3 years ago
7 0

Answer:

I =  26.36 cosω t A

Explanation:

Given that

C=0.74 mF

Vrms= 82 V

Frequency ,f= 49 Hz

We know that ω = 2 π f

ω = 2 x  π x 49

ω = 307.72 rad/s

As we know that voltage given as

V= Vo sinω t

V_o=\sqrt2\ V_{rms}

V_o=\sqrt2\ \times 82\

Vo=115.96 V

V=115.96 sinω t

The current given as

I=C\dfrac{dV}{dt}

I=0.74\times \dfrac{dV}{dt}\ mA

\dfrac{dV}{dt}=115.96\omega cos\omega t

I=0.74\times 115.96\times 307.22 cos\omega t\ mA

I = 26362.67 cosω t mA

I =  26.36 cosω t A

This is the current at time ant time t.

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A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
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Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

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then:

v = v₀e^(–t/τ)  

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v is the voltage after time t  

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t is time in seconds  

RC = τ = time constant  

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3 years ago
What are the characteristics of low energy waves?
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I think the correct answer from the choices presented above is the first option. The characteristics of low energy waves are <span>long wavelengths and low frequencies. Energy is established to be indirectly proportional to wavelengths and frequencies.</span>
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Now, let's plug in the variables we're given and simplify.

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3 years ago
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