Problem

A resistor of resistance $30 \, \Omega$ is connected in series with a capacitor of capacitance $79.5 \, \mu \mathrm{F}$ across a power supply of $50 \, \mathrm{Hz}$ and $100 \, \mathrm{V}$. Find:

1. The impedance ($Z$),
2. The current ($I$),
3. The phase angle ($\theta$),
4. The equation for the instantaneous value of current.

Data

• Capacitance: $C = 79.5 \, \mu \mathrm{F} = 79.5 \times 10^{-6} \, \mathrm{F}$
• Resistance: $R = 30 \, \Omega$
• Frequency: $f = 50 \, \mathrm{Hz}$
• RMS Voltage: $V_{\text{rms}} = 100 \, \mathrm{V}$

Prerequisite Concepts

1. Capacitive Reactance:

$$X_C = \frac{1}{2 \pi f C}$$

2. Impedance in an R-C Series Circuit:

$$Z = \sqrt{R^2 + X_C^2}$$

3. RMS Current:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$$

4. Phase Angle:

$$\theta = \tan^{-1} \left( \frac{X_C}{R} \right)$$

5. Instantaneous Current:

$$I_{\text{inst}} = I_{\text{max}} \sin \left( 2 \pi f t + \theta \right)$$

Where:

$$I_{\text{max}} = \sqrt{2} I_{\text{rms}}$$

Solution

Step 1: Calculate the Capacitive Reactance

$$X_C = \frac{1}{2 \pi f C}$$ $$X_C = \frac{1}{2 \times 3.14 \times 50 \, \mathrm{Hz} \times 79.5 \times 10^{-6} \, \mathrm{F}}$$ $$X_C = \frac{1}{24.963 \times 10^{-3}} = 40 \, \Omega$$

Step 2: Calculate the Impedance

$$Z = \sqrt{R^2 + X_C^2}$$ $$Z = \sqrt{(30 \, \Omega)^2 + (40 \, \Omega)^2}$$ $$Z = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$$

Step 3: Calculate the RMS Current

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$$ $$I_{\text{rms}} = \frac{100 \, \mathrm{V}}{50 \, \Omega}$$ $$I_{\text{rms}} = 2 \, \mathrm{A}$$

Step 4: Calculate the Phase Angle

$$\theta = \tan^{-1} \left( \frac{X_C}{R} \right)$$ $$\theta = \tan^{-1} \left( \frac{40 \, \Omega}{30 \, \Omega} \right)$$ $$\theta = 53.12^\circ$$

Step 5: Write the Equation for Instantaneous Current

$$I_{\text{inst}} = I_{\text{max}} \sin \left( 2 \pi f t + \theta \right)$$ $$I_{\text{max}} = \sqrt{2} I_{\text{rms}}$$ $$I_{\text{max}} = \sqrt{2} \times 2 \, \mathrm{A} = 2.828 \, \mathrm{A}$$ $$I_{\text{inst}} = 2.828 \, \mathrm{A} \times \sin \left( 2 \pi \times 50 \, t + 53.12^\circ \right)$$ $$I_{\text{inst}} = 2.828 \sin \left( 314 t + 53.12^\circ \right)$$

Answer

• Impedance: $Z = 50 \, \Omega$
• RMS Current: $I_{\text{rms}} = 2 \, \mathrm{A}$
• Phase Angle: $\theta = 53.12^\circ$
• Instantaneous Current: $I_{\text{inst}} = 2.828 \sin \left( 314 t + 53.12^\circ \right)$

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