In the astable multivibrator circuit shown in the figure, the frequency of oscillation at the output is
Show Hint
For the standard 555 astable circuit, memorize the frequency formula \(f \approx 1.44 / ((R_A + 2R_B)C)\). Remember that \(R_A\) and \(R_B\) are in ohms and C is in farads.
Step 1: Recall the frequency formula for a 555 timer in astable mode.
The frequency \(f\) is given by:
\[ f = \frac{1}{T} = \frac{1}{\ln(2) \cdot (R_A + 2R_B) \cdot C} \approx \frac{1.44}{(R_A + 2R_B)C} \]
Step 2: Identify the component values from the circuit diagram.
\( R_A = 7.5 \text{ k}\Omega = 7500 \, \Omega \)
\( R_B = 7.5 \text{ k}\Omega = 7500 \, \Omega \)
\( C = 0.1 \text{ }\mu F = 0.1 \times 10^{-6} \, F \)
Step 3: Substitute the values into the formula.
\[ f = \frac{1.44}{(7500 + 2 \cdot 7500) \cdot (0.1 \times 10^{-6})} \]
\[ f = \frac{1.44}{(7500 + 15000) \cdot (0.1 \times 10^{-6})} \]
\[ f = \frac{1.44}{22500 \cdot (0.1 \times 10^{-6})} = \frac{1.44}{2.25 \times 10^{-3}} \]
\[ f = \frac{1440}{2.25} = 640 \text{ Hz} \]
The calculated value is 640 Hz. The closest option is 644 Hz, which is likely the intended answer due to rounding of the constant 1.44 (more accurately 1/ln(2) \(\approx\) 1.443).
