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An AC circuit has an emf of 15.0cos(wt) V. There is also a 330. Ω resistor, a 5.10 nF capacitor a 0.0840 H inductor in the circuit. What is the value of the reactances XC and XL if the emf frequency is 2.30 kHz? XC = 1.36×104 Ω and XL = 1.21×103 Ω XC = 3.41×105 Ω and XL = 773 Ω XC = 2.71×104 Ω and XL = 2.43×103 Ω XC = 8.53×104 Ω and XL = 193 Ω

Question

An AC circuit has an emf of 15.0cos(wt) V. There is also a 330. Ω resistor, a 5.10 nF capacitor a 0.0840 H inductor in the circuit. What is the value of the reactances XC and XL if the emf frequency is 2.30 kHz? XC = 1.36×104 Ω and XL = 1.21×103 Ω XC = 3.41×105 Ω and XL = 773 Ω XC = 2.71×104 Ω and XL = 2.43×103 Ω XC = 8.53×104 Ω and XL = 193 Ω

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Solution

The reactance of a capacitor (Xc) and an inductor (XL) can be calculated using the following formulas:

Xc = 1/(2πfC) XL = 2πfL

where: f is the frequency, C is the capacitance, and L is the inductance.

Given: f = 2.30 kHz = 2300 Hz (since 1 kHz = 1000 Hz), C = 5.10 nF = 5.10 x 10^-9 F (since 1 nF = 10^-9 F), and L = 0.0840 H,

we can substitute these values into the formulas to find Xc and XL.

Xc = 1/(2π(2300)(5.10 x 10^-9)) ≈ 1.36 x 10^4 Ω XL = 2π(2300)(0.0840) ≈ 1.21 x 10^3 Ω

So, the reactances Xc and XL are approximately 1.36 x 10^4 Ω and 1.21 x 10^3 Ω, respectively.

This problem has been solved

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