
The source frequency for which a \[5\mu f\] capacitor has a reactance of $1000\Omega $.
A.$\dfrac{{100}}{\pi }Hz$
B.$\dfrac{{1000}}{\pi }Hz$
C.$200Hz$
D.$5000Hz$
Answer
476.1k+ views
Hint: The unit of capacitance reactance is the ohm. The capacitor is used to store the electrical energy.
They generally have internal resistance.
The capacitor doesn’t allow the DC to pass through it.
The capacitive reactance helps to measure the opposition to the Alternating current.
Complete answer:
Depending on the frequency of the applied current the capacitor also resists the flow of current. The greater reactance will give the smaller currents for the same voltage applied. We can compute the amplitude and phase changes using reactance.
The capacitance is measured in farads or microfarads.
Generally in the capacitors, the currents lead the voltage by $90$ degrees.
Given,
$C = 5\mu f$
We can solve the problem by using the formula below,
${X_1} = \dfrac{1}{{\omega L}}$
Now rearrange the above equation,
$\omega = \dfrac{1}{{{X_1}L}}$
Where,
$\omega $ is angular frequency
Now substitute the values in the above equation then it becomes,
$\omega = \dfrac{1}{{5 \times {{10}^{ - 6}}}} \times 1000$
Now solve the above equation we get the angular frequency is,
$\omega = 0.2 \times {10^3}$
From this, we can calculate the frequency,
We know,
$\omega = 2\pi f$
Then the frequency becomes,
$f = \dfrac{{2\pi }}{\omega }$
Now substitute the values in the above equation,
Then the frequency becomes,
$f = \dfrac{{100}}{\pi }cycles/\sec $
When we increase the frequency the capacitive reactance gets decreased. Also, the inductive reactance will be increased.
Finally, the correct answer is an option (A).
Note:
Capacitors are also used to suppress the undesirable frequency and they are sometimes called the filter capacitor.
The property of capacitive reactance is ideally used in the AC filter circuits.
It is also used to avoid unwanted ripple voltages and short circuit signal paths.
Signal coupling and signal decoupling are some applications of capacitors.
They generally have internal resistance.
The capacitor doesn’t allow the DC to pass through it.
The capacitive reactance helps to measure the opposition to the Alternating current.
Complete answer:
Depending on the frequency of the applied current the capacitor also resists the flow of current. The greater reactance will give the smaller currents for the same voltage applied. We can compute the amplitude and phase changes using reactance.
The capacitance is measured in farads or microfarads.
Generally in the capacitors, the currents lead the voltage by $90$ degrees.
Given,
$C = 5\mu f$
We can solve the problem by using the formula below,
${X_1} = \dfrac{1}{{\omega L}}$
Now rearrange the above equation,
$\omega = \dfrac{1}{{{X_1}L}}$
Where,
$\omega $ is angular frequency
Now substitute the values in the above equation then it becomes,
$\omega = \dfrac{1}{{5 \times {{10}^{ - 6}}}} \times 1000$
Now solve the above equation we get the angular frequency is,
$\omega = 0.2 \times {10^3}$
From this, we can calculate the frequency,
We know,
$\omega = 2\pi f$
Then the frequency becomes,
$f = \dfrac{{2\pi }}{\omega }$
Now substitute the values in the above equation,
Then the frequency becomes,
$f = \dfrac{{100}}{\pi }cycles/\sec $
When we increase the frequency the capacitive reactance gets decreased. Also, the inductive reactance will be increased.
Finally, the correct answer is an option (A).
Note:
Capacitors are also used to suppress the undesirable frequency and they are sometimes called the filter capacitor.
The property of capacitive reactance is ideally used in the AC filter circuits.
It is also used to avoid unwanted ripple voltages and short circuit signal paths.
Signal coupling and signal decoupling are some applications of capacitors.
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