Question

In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ε₀ = 50 V and ν = 50/π Hz. Find the peak current and the average power dissipated in the circuit.

Answer 1

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε₀ = 50 V
Frequency of the AC source, ν = 50/$\mathrm{\pi }$ Hz
Capacitive reactance $\left(X_C\right)$ is given by,
$X_C=\frac{1}{\omega C}$
Here, $\omega$ = angular frequency of AC source
C = capacitive reactance of capacitance
$$\begin{gathered} \therefore X_C=\frac{1}{2\mathrm{\pi }\times {\displaystyle \frac{50}{\mathrm{\pi }}}\times 25\times {10}^{-6}} \\ \Rightarrow X_C=\frac{{10}^4}{25}\mathrm{\Omega } \end{gathered}$$
Net reactance of the series RC circuit $\left(Z\right)$ = $\sqrt{R^2+{\left(X_C\right)}^2}$
$\Rightarrow$ Z = $\sqrt{{\left(300\right)}^2+{\left(\frac{{10}^4}{25}\right)}^2}$
= $\sqrt{{\left(300\right)}^2+{\left(400\right)}^2}$ = 500 $\mathrm{\Omega }$
(a) Peak value of current $\left(I_0\right)$ is given by,
$$\begin{gathered} I_0=\frac{{\epsilon }_0}{Z} \\ \Rightarrow I_0=\frac{50}{500}=0.1\mathrm{A} \end{gathered}$$

(b) Average power dissipated in the circuit $\left(P\right)$ is given by,
P = ${\epsilon }_{\mathrm{rms}}$I_rms cosϕ.
${\epsilon }_{rms}$ = $\frac{{\epsilon }_0}{\sqrt{2}}$
and $I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}$
$$\begin{gathered} \therefore P=\frac{E_0}{\sqrt{2}}\times \frac{I_0}{\sqrt{2}}\times \frac{R}{Z} \\ \Rightarrow P=\frac{50\times 0.1\times 300}{2\times 500} \\ \Rightarrow P=\frac{3}{2}=1.5\mathrm{W} \end{gathered}$$