Finding XL-XC in RLC circuit with 6.72A current and 55.9° phase angle

[waywardtigerlily]

In a certain RLC circuit, the rms current is 6.72A, the rms voltage is 230V, and the current leads the voltage by 55.9 Degrees.
a)what is the resistance of the circuit?
b)calculate the total XL-Xc
c)Calc. the average power dissipated in the circuit.

a) 1.92E1 Ohms
b)?
c)8.665E2 J

I can't figure out b! I tried using

tan x = X/R
tan 55.9 = (Xl-Xc)/(1.92E1 Ohm)
But this did not give me the right answer...maybe i am just doing the math wrong here...would you do this to solve for (XL-Xc)?
(1.92E1 Ohm)tan 55.9 = "x" ?
Thanks

 

[Andrew Mason]

In a certain RLC circuit, the rms current is 6.72A, the rms voltage is 230V, and the current leads the voltage by 55.9 Degrees.
a)what is the resistance of the circuit?
b)calculate the total XL-Xc
c)Calc. the average power dissipated in the circuit.

Your approach is right.

I get 28.36 ohms.

Note: The power is in units of energy / time. $VIcos\phi = P = I^2R$, so it is 867 J/sec.

AM

 

[thepatientmental]

for your help!

a) The resistance of the circuit can be calculated using Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I). Therefore, R = 230V / 6.72A = 1.92E1 Ohms.

b) To calculate the total XL-Xc, we need to know the individual values of XL and XC. XL is the inductive reactance, which is calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. XC is the capacitive reactance, which is calculated using the formula XC = 1/(2πfC), where C is the capacitance. Since we do not have these values, we cannot calculate the total XL-Xc.

c) The average power dissipated in the circuit can be calculated using the formula P = VIcosθ, where V is the voltage, I is the current, and θ is the phase angle between them. In this case, V = 230V, I = 6.72A, and θ = 55.9 degrees. Converting θ to radians, we get θ = 0.976 radians. Therefore, P = (230V)(6.72A)cos(0.976) = 8.665E2 J.