# AC Circuits

## AC Circuits

#### Purpose

To build a low pass filter using a capacitor and resistor in series and to investigate how the filter response as a function of the frequency of the input signal.

#### Objectives

• To learn how to use of a digital oscilloscope
• To calculate the impedance of a AC circuit or a resistor and capacitor in series
• To learn the relation between the voltage across each element in an AC circuit of R and C element in series.
• To find out the dependence of the voltage across the C element as a function of the frequency of the input signal

#### Equipment

• Digital Oscilloscope Tektronix TDS 2001C
• Function Generator
• 3 BNC cables, 1 banana-banana plug cable

#### Definitions

A capacitor in an AC circuit has capacitive reactance of:

$X_{C}=\frac{1}{\omega C} = \frac{1}{2\pi f C}$

An inductor in an AC circuit has inductive reactance of:

$X_{L} = \omega L = 2 \pi f L$

#### Activity 1. RC Low Pass Filter

###### Theory

We built a low-pass filter by connecting a resistor and capacitor in series. The impedance of the circuit is:

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

If the current in the circuit is given by $I=I_o \sin \omega t$, the maximum voltage across is:

$V_o=ZI_o$

The maximum voltage across the capacitor is:

$V_{oC}=X_C I_o$

The ratio between the two then is given as:

$\frac{V_{oC}}{V_o} = \frac{X_C}{Z}=\frac{1}{2\pi f C \sqrt{R^2 + 4 \pi^2 f^2 C^2}}$

or

$\frac{V_{oC}}{V_o} = \frac{1}{\sqrt{4\pi^2 f^2 C^2 R^2+1}}$

When the frequency is zero the ratio is simply 1. When the frequency becomes very large ($f \rightarrow \infty$), the limit of the ratio goes to zero. The cut-off frequency is that frequency for which the ratio is one half:

$\frac{1}{2}=\frac{1}{\sqrt{4\pi^2 f^2 C^2 R^2+1}} \rightarrow f = \frac{1}{2\pi RC}$

###### Experimental Setting

Use a 1.0-kΩ resistor and a 0.1-μF capacitor to connect them in series to the Function Generator and and turn the power on.