POWER IN AN L-R-C SERIES CIRCUIT For the L-R-C series circuit of Example 31.4, (a) calculate the power factor and (b) calculate the average power delivered to the entire circuit and to each circuit element.

Question

POWER IN AN L-R-C SERIES CIRCUIT

For the [latex]L-R-C[/latex] series circuit of Example 31.4, (a) calculate the power factor and (b) calculate the average power delivered to the entire circuit and to each circuit element.

Accepted Answer

IDENTIFY and SET UP:

We can use the results of Example 31.4.
The power factor is the cosine of the phase angle [latex]\phi[/latex], and we use Eq. (31.31) to find the average power delivered in terms of [latex]\phi[/latex] and the amplitudes of voltage and current.

[latex]P_{\mathrm{av}}=\frac{1}{2} V I \cos \phi=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi[/latex] (31.31)

EXECUTE:

(a) The power factor is cos [latex]\phi[/latex] = cos 53° = 0.60.
(b) From Eq. (31.31),

[latex]P_{\mathrm{av}}=\frac{1}{2} V I \cos \phi=\frac{1}{2}(50 \mathrm{~V})(0.10 \mathrm{~A})(0.60)=1.5 \mathrm{~W}[/latex]

EVALUATE: Although [latex]P_{av}[/latex] is the average power delivered to the [latex]L-R-C[/latex] combination, all of this power is dissipated in the resistor.

As Figs. 31.16b and 31.16c show, the average power delivered to a pure inductor or pure capacitor is always zero.

Question 31.7: POWER IN AN L-R-C SERIES CIRCUIT For the L-R-C series circui...

Related Answered Questions