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Cloud [144]
2 years ago
8

For the RC circuit and the RL circuit, assume that the period of the source square wave is much larger than the time constant fo

r each. Make a sketch of vR(t) as a function of t for each of the circuits?

Physics
1 answer:
patriot [66]2 years ago
7 0

Answer with Explanation:

Concepts and reason

The concept to solve this problem is that if a capacitor is connected in a RC circuit then it allows the flow of charge through circuit only till it gets fully charged. Once the capacitor is charged it will not allow any charge or current to flow.

Opposite is the case with inductor in the RL circuit. According to Faraday's law an inductor develops an emf to oppose the voltage applied but once the flux change stops then the inductor behaves just like a normal wire as if no inductor is there.

In attached figure, resistor is connected in series to the capacitor.

As we considered V_{C} the voltage across the capacitor and V_{s} the voltage across the source.

Voltage across a resistor In RC circuit.

V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )

Voltage across a resistor In RL circuit.

V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )

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Answer:x=2 and x=3

Explanation:

Given

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U(x)=2x^3-15x^2+36x-23

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For Force to be zero F=0

\Rightarrow 6x^2-30x+36=0

\Rightarrow x^2-5x+6=0

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Explanation:
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Answer:

  E =  k Q    1 / (x₀-x₂) (x₀-x₁)

Explanation:

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In this case as we have a continuous load distribution we can use the concept of linear density

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We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

           E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

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We replace the density

             E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

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Answer:

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Explanation:

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then: part(a):

\text{heat loss } =  \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

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\text {heat loss} = h_1 (T_1 - T_{s1})

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