How to calculate capacitance given measured voltage,frequency and resistance?

I believe you meant capacitive reactance is 47 ohms. If that is the case, your answer is 5.69231042494974E-07 farads. Below is the the relevant calculation method in my handy dandy VB program ElseIf F > 0 Then If XC > 0 Then C = 1 / (2 * Pi * F * XC) This calculation was performed using double precision numbers on a computer. Where F = frequency XC = Capacitive reactance C = capacitance in farads Pi = 4*Atn(1) or 3.1416......... Total time for calculation wasless than 1 second. Final Edit I found the error you mentioned. My formula was correct, but in the area where the form is booted, there was a statement Pi=4*Tan(1) which is indeed 2 *Pi. This statement obviously overrode the Pi=4*Atn(1). My application is now corrected. Thanks. The correct answer is, as you said, 1.12875846164465E-06 Farads. I stand corrected. The last version of my program was an update to include more calculations.

You left out a given. I don`t think there is enough information given to calculate the capacitance value. Total current or the Voltage across the 47 Ohm resistor would help. Also need to know if the capacitor and resistor are connected in series or parallel. Edit addition after receiving your additional information. Apparently the 47 Ohms is the capacitive reactance of the capacitor rather than the value of a resistor in the circuit. If so it should be called reactance (Xc) instead of resistance even though reactance and resistance are both expressed in Ohms. Assuming Xc = 47 Ohms then: Xc = 1 / 2pi fC 47 Ohm = 1/[(6.28) X (3,000 Hz) X (C)] C = 1 / [(6.28) X (3,000 Hz) X (47 Ohm)] C = .000001129 Farads = approximately 1 uF 2nd edit addition: After seeing "past EE texasmav`s" answer that was calculated in less than 1 second on his handy dandy VB program using double precision numbers on a computer I realized something was amiss. Low and behold it looks like some one has inadvertently used 4pi instead of 2pi as one of the inputs to the handy dandy VB program. I could be wrong of course. If so I feel relatively certain that my mistake will be caught by the keen eyes of one of the other responders and corrected. Chuckle, chuckle.

In this case, we have a formula to calculate the frequency..... ==> f = 1 / (2*pi*R*C) Using this formula, we have C = 1 / ( 2*pi*f*R ) [Voltage is not required] So, C = 1/ (2*3.14*3000*47) ==>C = 0.0000011 F ==>C = 1.1 micro F >=====================================< ANSWER Hope this will help you...........

You left out a given. I don`t think there is enough information given to calculate the capacitance value. Total current or the Voltage across the 47 Ohm resistor would help. Also need to know if the capacitor and resistor are connected in series or parallel. Edit addition after receiving your additional information. Apparently the 47 Ohms is the capacitive reactance of the capacitor rather than the value of a resistor in the circuit. If so it should be called reactance (Xc) instead of resistance even though reactance and resistance are both expressed in Ohms. Assuming Xc = 47 Ohms then: Xc = 1 / 2pi fC 47 Ohm = 1/[(6.28) X (3,000 Hz) X (C)] C = 1 / [(6.28) X (3,000 Hz) X (47 Ohm)] C = .000001129 Farads = approximately 1 uF 2nd edit addition: After seeing "past EE texasmav`s" answer that was calculated in less than 1 second on his handy dandy VB program using double precision numbers on a computer I realized something was amiss. Low and behold it looks like some one has inadvertently used 4pi instead of 2pi as one of the inputs to the handy dandy VB program. I could be wrong of course. If so I feel relatively certain that my mistake will be caught by the keen eyes of one of the other responders and corrected. Chuckle, chuckle.

As for my understanding of electrical engineering is concerned, my approach for the experiment would have been very slightly different. I would have checked the value of current also, when voltage is applied across the inductor. Also if the inductor was with air-core, or with ferromagnetic-core. In case of ferromagnetic core, I would have measured the wattage consumed also. This is because of the following reasons:- (a) In case of air core inductor, the measured resistance and the actual AC resistances are expected to be almost same. That is Zr = 47 ohm (b) In case of ferromagnetic core inductor, the AC resistance should represent the Iron Loss also. Hence there would have been a different value than the measured resistance. In that case the AC resistance could be calculated as (Iac x Iac) x Zr = W Now with regard to the steps of calculations:- Zac = Equivalent Impedance Zr = AC resistance Zl = Impedance due to inductance. Vac = Terminal voltage =3.97 v Iac = to be given Iac x Z ac = Vac = 3.97 v So Zac = V ac / Iac Zac = √ [(Zl x Zl) +(ZrxZr)] Having known the value of Zac, and Zr from the above formula we can find the value of Zl. And so we can find the value of L from the formulas given in the replies of the earlier responders

In this case, we have a formula to calculate the frequency..... ==> f = 1 / (2*pi*R*C) Using this formula, we have C = 1 / ( 2*pi*f*R ) [Voltage is not required] So, C = 1/ (2*3.14*3000*47) ==>C = 0.0000011 F ==>C = 1.1 micro F >=====================================< ANSWER Hope this will help you...........