17.10 AC Through RC-Series Circuit

17.10 AC Through RC-Series Circuit

Consider a network of resistance $R$ and a capacitor $C$ connected in series by an alternating voltage source $V$ as shown in Fig. 17.9 (a). As $R$ and $C$ are in series, so same current would flow through each of them. The potential difference $V$ across the resistance $R$ would be $I_{\mathrm{rms}} R$ and it would be in phase with the current. The vector diagram of the voltage and current is shown in Fig. 17.9 (b). Taking the current as reference, potential difference $V_{R} = I_{\mathrm{rms}} R$ across the resistance is represented by a line along the current line because potential drop $I_{\mathrm{rms}} R$ is in phase with current. The potential difference across the capacitor will be: $V_{C} = I_{\mathrm{rms}} X_{C} = \dfrac{I_{\mathrm{rms}}}{\omega C}$ As this, voltage lags behinds the current by $\pi/2$ rad or $90^{\circ}$, so the line representing the vector $I_{\mathrm{rms}} X_{C}$ is drawn at right angles to the current line Fig. 17.9 (b).

(Fig. 17.9(a))

The applied voltage $V_{\mathrm{rms}}$ that will send the current $I$ in the circuit is obtained by resultant of the vectors $I_{\mathrm{rms}} R$ and $\dfrac{I_{\mathrm{rms}}}{\omega C}$ i.e; $V_{\mathrm{rms}} = \sqrt{(V_{R})^{2} + (V_{C})^{2}}$ $V_{\mathrm{rms}} = \sqrt{(I_{\mathrm{rms}} R)^{2} + \left(\dfrac{I_{\mathrm{rms}}}{\omega C}\right)^{2}}$ $V_{\mathrm{rms}} = I_{\mathrm{rms}}\sqrt{R^{2} + \left(\dfrac{1}{\omega C}\right)^{2}}$ $V_{\mathrm{rms}} = I_{\mathrm{rms}}\sqrt{R^{2} + \left(\dfrac{1}{\omega C}\right)^{2}}$ Impedance

(Fig. 17.9(b))

Equation 17.21 suggests that we can find the impedance of a series AC circuit by vector addition. The resistance $R$ is represented by a horizontal line in the direction of current which is taken as reference. The reactance $X_{C} = 1 / \omega C$ is shown by a line lagging the $R$-$Q$ line by $90^{\circ}$ as shown in Fig. 17.9(c). The impedance $Z$ of the circuit is obtained by the vector summation of resistance and reactance. Figure 17.9(c) is known as impedance diagram of the circuit. The angle which the line representing the impedance $Z$ makes with $R$ line gives the phase difference between the voltage and current $I$ in Fig. 17.9(c), the current is leading the voltage applied by the angle $\theta$ as shown in Fig. 17.9(d) and is given by $\theta = \tan^{-1}\left(\dfrac{X_{C}}{R}\right) = \tan^{-1}\left(\dfrac{1}{\omega CR}\right) = \tan^{-1}\left(\dfrac{1}{2\pi f C R}\right)$ The power consumed in RC-series circuit is primarily due to the resistor, as the capacitor only stores energy and does not dissipate it. The power consumed by the resistor can be calculated by using the formula: $P = I^{2} R$ or

where $I$ is the current, $R$ is the resistance, $V$ is the voltage, and $\theta$ is the phase angle between voltage and current.

(Fig. 17.9(c): Impedance diagram of RC series circuit.)

(Fig. 17.9(d): Voltage and current waveform in RC-series circuit.)

Example 17.6

A resistor of resistance $5\,\mathrm{k}\Omega$ is connected in series with capacitor of capacitance $5\,\mathrm{F}$ across the AC source of $220\,\mathrm{V}$ that has frequency $50\,\mathrm{Hz}$. Calculate the phase angle.

Solution

Resistance $R = 5\,\mathrm{k}\Omega = 5000\,\Omega$ , Capacitance $C = 5\,\mu\mathrm{F} = 5 \times 10^{-6}\,\mathrm{F}$ and Frequency $f = 50\,\mathrm{Hz}$ We know that; Phase angle $\theta = \tan^{-1}\left(\dfrac{1}{\omega CR}\right)$ As $\omega = 2\pi f$ , so $\theta = \tan^{-1}\left(\dfrac{1}{2\pi f CR}\right)$ Putting the values $\theta = \tan^{-1}\left(\dfrac{1}{2 \times 3.14 \times 50\,\mathrm{Hz} \times 0.000005\,\mathrm{F} \times 5000\,\mathrm{W}}\right)$ $\theta = \tan^{-1}(0.1273) = 7.25^{\circ}$