Question

In an L-R-C series circuit, the components have the following values: L = 20.0 mH, C = 140 nF, and R = 350Ω. The generator has an rms voltage of 120V and a frequency of 1.25 kHz. Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

Short answer

The power supplied by the generator is 7.32 W and the power dissipated by the generator is 7.32 W.

Step-by-step solution

Step-1: Formula of power supplied and power dissipated  

$\mathit{\varphi }$is the phase difference between voltage and current.

$\mathbf{\varphi }\mathbf{}\mathbf{is}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{wL}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{\mathbf{\omega C}}}}{\mathbf{R}}\mathbf{\right)}$where$\varphi$ is the phase difference between voltage and current,$\omega$ is the angular frequency of voltage supply, L is the value of inductance of the inductor, C is the capacitance, R is the resistance

${\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$represents the d.c. current/voltage that dissipates the same amount of power as the average power dissipated by the alternating current/voltage.

${\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{r}\mathbf{m}\mathbf{s}}}{\mathbf{Z}}$

${\mathrm{P}}_{\mathrm{avg}}$is the average power dissipated by the circuit.

${\mathbf{P}}_{\mathbf{avg}}\mathbf{=}{\mathbf{V}}_{\mathbf{rms}}{\mathbf{I}}_{\mathbf{rms}}\mathbf{cos\varphi }\mathbf{=}{\left(I_{\mathrm{rms}}\right)}^{\mathbf{2}}\mathbf{R}$

$V_{rms}$and$I_{rms}$ are the rms voltage and current respectively.

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$\mathit{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}\mathbf{(}{\mathbf{X}}_{\mathbf{L}}\mathbf{-}{\mathbf{X}}_{\mathbf{C}}\mathbf{C}{\mathbf{)}}^{\mathbf{2}}}$, where Z is the impedance

Step-2: Calculations for power supplied by the generator

$\varphi =\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{157\Omega -909\Omega }{350\Omega }\right)-65.04°$

$\mathit{z}\mathbf{=}\sqrt{{350}^2+{\left(157-909\right)}^2}\Omega =829.8\Omega$

$I_{rms}=\frac{120V}{829.8\Omega }=0.145A$

Finally,${\mathrm{P}}_{\mathrm{avg}}=(120.\mathrm{V})\left(0145\mathrm{A}\right)\cos (-65.04)=7.32\mathrm{W}$

Therefore the value of the power supplied by the generator is 7.32W

Step-3: Calculations for power dissipated by the resistor

$P_{avg}=(0.145A{)}^2(350\Omega )=7.32W$

Therefore the value of the power supplied by the generator is 7.32W and the value of the power dissipated by the generator is 7.32W.