Verification of the capacitance of a capacitor

AI Thread Summary
To verify the capacitance of a capacitor that results in a phase difference of 15 degrees in an AC circuit, the relationship between phase difference, capacitive reactance, and resistance must be applied. The equation tan θ = -Xc/R indicates that as capacitance increases, capacitive reactance decreases, leading to a smaller phase difference. The discussion confirms that a higher capacitance value is necessary to achieve the desired phase shift. It is noted that the effect of the capacitor diminishes with increased capacitance, resulting in a reduced phase difference between current and voltage. Understanding these principles is essential for accurate capacitance verification in circuit analysis.
Samar A
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Homework Statement


In the circuit shown in the figure below, the hase difference between the current and the voltage is 30 degrees, show how could you verify the capacitance of the conductor so that the phase difference could be 15 degrees.
Muz9U.jpg

Homework Equations


tan θ=-Xc/R (the phase difference in the RC cirucit with an AC source)

The Attempt at a Solution


That is my attempt, I show how much would be the value of the capacitance in the second case mathematically.
2VX1D.png

Is it correct?
 
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Samar A said:
show how could you verify the conductance **capacitance** of the conductor so that the phase difference could be 15 degrees.
Conductance is the reciprocal of resistance. What you have in the circuit is a capacitor (or condenser).
I didn't check your actual values step by step, but your approach is correct. Also, the new capacitance is coming out to be more than the old capacitance, which is correct (can you think why?).
 
cnh1995 said:
Conductance is the reciprocal of resistance. What you have in the circuit is a capacitor (or condenser).
Oh, sorry, I corrected it.
cnh1995 said:
can you think why?).
OK, I think because when the capacitance is greater the reactance of the capacitor will be less than the original one, so its effect in the circuit will be smaller, and then the phase difference between the current and the voltage will be smaller too. Am I right?
 
Samar A said:
OK, I think because when the capacitance is greater the inductance **reactance** of the capacitor will be less than the original one, so its effect in the circuit will be smaller, and then the phase difference between the current and the voltage will be smaller too. Am I right?
Right. Capacitive reactance is inversely proportional to the capacitance.
 
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