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How do you write out an equation used to find the inductive reactance ant the capacitive reactance?
Wiki User
∙ 14y ago
Xc(capacitive reactance) = 1/(2piFC)
XL(inductive reactance) = 2piFL
Where pi=3.14etc.,
F=frequency and C and L are capacitance and inductance.
Please pardon lack of proper symbology.
Wiki User
∙ 14y ago
Wiki User
∙ 14y agoInductive reactance, XL=2piifl, where f is frequency and l is the inductance.
Capacitive reactance, XC=1/2piifc, where f is frequency and c is the capacitance.
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Write the relationship between Impedance Resistance and Reactance using Operator J?
The operator, 'j', is used to indicate a phasor quantity that has been rotated, counterclockwise, through an angle of 90 degrees.So, if (for example) the operator is applied to a voltage U, then it is written as jU, which indicates means that the voltage lies along the vertical positive axis. A further operation by j, results in jjU, or j2 U, which means that the voltage lies along the horizontal negative axis -so, j2 is equivalent to -1U (or j is equivalent to the square-root of -1) or, simply, -U.A further operation by j, results in the voltage lying along the negative vertical axis: that is: jjjU = jj2U=-jU.But to answer your question, for inductive reactance (XL), we express impedance (Z) a follows: Z = R+jXL and, for capacitive reactance (XC), we express impedance as Z = R - jXC (the L and C should be subscripts).Strictly-speaking the operator j doesn't actually apply to impedance, because impedance, resistance, and reactance or not phasor (vector) quantities. So if we wanted to be strictly accurate, the above equations should be written as:(E/I) = (UR/I) + j (UL/I) and (E/I) = (UR/I)- j (UC/I)...but this is being rather pedantic.
Why does A.C. not follow OHM's law?
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Reactance of capacitor is inversely proportional to frequency. I should not need to write the exact equation here, its in your textbook. All you need is that its inversely proportional to frequency for proof.We will now assume an ideal capacitor to keep the analysis simple.at DC the frequency is zero, the inverse of this is infinite reactance: open circuitat low frequency AC frequency is low, the inverse of this is high reactanceat midrange frequency AC frequency is midrange, the inverse of this is midrange reactanceat high frequency AC frequency is high, the inverse of this is low reactanceat infinite frequency AC frequency is infinite, the inverse of this is zero reactance: short circuitThis disproves your original statement as written, except for the special cases of DC and infinite frequency AC (which does not occur), for ideal capacitors.As all real capacitors are nonideal, they have leakage resistance. This means that even for the special case of DC the capacitor is not a true open circuit, just a very high resistance resistor. Which also disproves it for the remaining case of DC in real capacitors.
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What is total impedance of a circuit when the resistance is 15k ohms and the capacitive reactance is 10k ohms?
You can write this as a complex number; the resistance is the real part, the reactance is the imaginary part (negative, for a capacitive reactance): 15 + j10 kilohms. ("j" is used instead of "i", to avoid confusion with current, which is symbolized by "i".) This is in rectangular coordinates; with a scientific calculator you can use rectangular --> polar conversion, to get the absolute value and the angle. To get just the absolute value, use Pythagoras' Theorem, which in this case gives about 18 kilohms.You can write this as a complex number; the resistance is the real part, the reactance is the imaginary part (negative, for a capacitive reactance): 15 + j10 kilohms. ("j" is used instead of "i", to avoid confusion with current, which is symbolized by "i".) This is in rectangular coordinates; with a scientific calculator you can use rectangular --> polar conversion, to get the absolute value and the angle. To get just the absolute value, use Pythagoras' Theorem, which in this case gives about 18 kilohms.You can write this as a complex number; the resistance is the real part, the reactance is the imaginary part (negative, for a capacitive reactance): 15 + j10 kilohms. ("j" is used instead of "i", to avoid confusion with current, which is symbolized by "i".) This is in rectangular coordinates; with a scientific calculator you can use rectangular --> polar conversion, to get the absolute value and the angle. To get just the absolute value, use Pythagoras' Theorem, which in this case gives about 18 kilohms.You can write this as a complex number; the resistance is the real part, the reactance is the imaginary part (negative, for a capacitive reactance): 15 + j10 kilohms. ("j" is used instead of "i", to avoid confusion with current, which is symbolized by "i".) This is in rectangular coordinates; with a scientific calculator you can use rectangular --> polar conversion, to get the absolute value and the angle. To get just the absolute value, use Pythagoras' Theorem, which in this case gives about 18 kilohms.
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Write the relationship between Impedance Resistance and Reactance using Operator J?
The operator, 'j', is used to indicate a phasor quantity that has been rotated, counterclockwise, through an angle of 90 degrees.So, if (for example) the operator is applied to a voltage U, then it is written as jU, which indicates means that the voltage lies along the vertical positive axis. A further operation by j, results in jjU, or j2 U, which means that the voltage lies along the horizontal negative axis -so, j2 is equivalent to -1U (or j is equivalent to the square-root of -1) or, simply, -U.A further operation by j, results in the voltage lying along the negative vertical axis: that is: jjjU = jj2U=-jU.But to answer your question, for inductive reactance (XL), we express impedance (Z) a follows: Z = R+jXL and, for capacitive reactance (XC), we express impedance as Z = R - jXC (the L and C should be subscripts).Strictly-speaking the operator j doesn't actually apply to impedance, because impedance, resistance, and reactance or not phasor (vector) quantities. So if we wanted to be strictly accurate, the above equations should be written as:(E/I) = (UR/I) + j (UL/I) and (E/I) = (UR/I)- j (UC/I)...but this is being rather pedantic.
