For the following diagram, RA = 2.2K Ohm, RB = 3.9K Ohm, and C = 0.1 micro-Farads. Determine the free-running frequency (KHz).

1. Break Down the Problem

We need to calculate the free-running frequency of an astable multivibrator circuit, which typically involves resistors RA, RB, and capacitor C. The formula for the free-running frequency $f$ in kilohertz (KHz) is given by:

$f = \frac{1.44}{(RA + 2RB) \times C}$

2. Relevant Concepts

• Resistors: RA and RB act to set the timing for the multivibrator.
• Capacitance: C determines how quickly the circuit can charge and discharge.
• Frequency Calculation: We will use the given formula.

3. Analysis and Detail

1. Convert the resistances and capacitance to standard units:

   • $RA = 2.2K \: \Omega = 2200 \: \Omega$
   • $RB = 3.9K \: \Omega = 3900 \: \Omega$
   • $C = 0.1 \: \mu F = 0.1 \times 10^{-6} \: F = 1 \times 10^{-7} \: F$

2. Substitute values into the frequency formula:

$f = \frac{1.44}{(2200 + 2 \times 3900) \times (1 \times 10^{-7})}$

3. Calculate the total resistance $RA + 2RB$: $RA + 2RB = 2200 + 2 \times 3900 = 2200 + 7800 = 10000 \: \Omega$

4. Plug into the frequency formula: $f = \frac{1.44}{10000 \times 1 \times 10^{-7}} = \frac{1.44}{1 \times 10^{-3}} = 1440 \: Hz$

5. Convert to kilohertz: $f = \frac{1440}{1000} = 1.44 \: KHz$

4. Verify and Summarize

The calculations have been verified for accuracy:

• Calculated $RA + 2RB = 10000 \: \Omega$.
• Substituted into the frequency formula yielding $1.44 \: KHz$.

Final Answer

The free-running frequency is 1.44 KHz.