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9 months ago

7

In a series rlc ac circuit, a second resistor is connected in parallel with the resistor previously in the circuit. as a result

of this change, what happens to the impedance of the circuit?

1 answer:

raketka [301]9 months ago

6 0

According to the given statement:

The frequency response does not change, which is the first thing we notice.

The new resistance at the resonance point causes a reduction in the circuit's current flow.
Z = R + R₂

<h3>The definition of series circuits:</h3>

electrical circuit. The path that the entire current takes as it passes through each component makes up a series circuit. Branching is used in parallel circuits to divide the current and limit the amount that flows through each branch.

<h3>How does a series circuit operate?</h3>

According to this definition, there are three principles of series circuits: all parts share the same current, resistances add up to a larger total resistance, and voltage drops add up to a larger total voltage. In the definition of a series circuit, all of these guidelines have their origin.

<h3>According to the given information:</h3>

The impedance of a series circuit is

Z₀² = R² + (X $_L$ -X $_C$ ) ²

The initial resistance impedance shifts to when we add another resistor to the series

Z² = (R + R₂) ² + (X $_L$ - X $_C$ ) ²

Let's examine this sentence.

The frequency response remains unchanged, which is the first thing we notice.
The new resistance at the resonance point causes the circuit's current to decrease.

Z = R + R₂

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The angular velocity is $w = 53.35 \ rounds /minute$

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From the question we are told that

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The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

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Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

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Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

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$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

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This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

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Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

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So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

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