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yarga [219]
4 years ago
15

For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

rcuit? the current leads the generator EMF by 0.982 radians, what are IT 3 20 the resistance value of the resistor, R?
Physics
1 answer:
MAXImum [283]4 years ago
5 0

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

z = \frac{V_o}{i_o}

now we will have

z = \frac{125}{3.20} = 39.06 ohm

Part b)

Now we also know that

\frac{R}{z} = cos\theta

\theta = 0.982 rad = 56.3 degree

now we have

\frac{R}{39.06} = cos56.3

R = 21.7 ohm

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<h3>Explanation</h3>

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\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0.

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  • v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2},
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\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}

when the ball reaches the goal.

\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x.

Solve this quadratic equation for x, x > 0.

  • x = 65.1 meters when h = 0 meters.
  • x = 6.54 or 58.5 meters when h = 4 meters.

In other words,

  • For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
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