Question

In the astable multivibrator circuit shown in the figure, the frequency of oscillation at the output is

• A. 544 Hz
• B. 54.4 Hz
• C. 64.4 Hz
• D. 644 Hz

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.

Answer 1

The Correct Option is D

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).