An RLC series circuit contains a capacitor with a capacitance of 66.3 μF, an inductor with an inductance of 0.0663 H, and a resistor with a value of 8 Ω connected to a 120-V, 60-Hz line (Figure 24–7). How much current will flow in this circuit?

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Accepted Answer

The first step in solving this problem is to find the values of capacitive and inductive reactance:

[latex]X_{C} = \frac{1}{ 2 \pi f C}[/latex]

[latex]X_{C} = \frac{1}{377 imes 0.0000663 F} [/latex]

[latex]X_{C} = 40.008 \Omega[/latex]

[latex]X_{L} = 2 \pi f L[/latex]

[latex]X_{L} = 377 imes 0.0663 H [/latex]

[latex]X_{L} = 24.995 \Omega[/latex]

Now that the capacitive and inductive reactance values are known, the circuit impedance can be found using the formula

[latex]Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}[/latex]

[latex]Z = \sqrt{(8 \Omega)^{2} + (24.995 \Omega - 40.008 \Omega)^{2}}[/latex]

[latex]Z = \sqrt{64 \Omega + 225.39 \Omega}[/latex]

[latex]Z = 17.011 \Omega[/latex]

Now that the circuit impedance is known, the current flow can be found using Ohm’s law

[latex]I_{T} = \frac{E_{T}}{Z}[/latex]

[latex]I_{T} = \frac{120 V}{17.011 \Omega}[/latex]

[latex]I_{T} = 7.054 A[/latex]

Question 24.2: An RLC series circuit contains a capacitor with a capacitanc...

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