Questions: Oscillating frequency relation with R and C of the given circuit. Set the representative time constant of the circuit. Given that Rs = 470 kΩ, the lower resistance value of the first capacitor can be taken as mind of the test class. Separate your answer with commas.
Transcript text: Oscillating frequency relation with R and C of the given circuit. Set the representative time constant of the circuit. Given that Rs = 470 kΩ, the lower resistance value of the first capacitor can be taken as mind of the test class. Separate your answer with commas.
Solution
Solution Steps
Step 1: Understanding the Problem
The problem involves determining the oscillating frequency relation with resistance \( R \) and capacitance \( C \) in a given circuit. Additionally, we need to set the representative time constant of the circuit. The problem provides a list of numbers and specifies that \( R_s = 470 \, \text{k}\Omega \). We are also asked to identify the lower resistance value of the first capacitor, which is described as the "mind of the test class."
Step 2: Identifying the Relevant Values
The list of numbers provided is: 7.30, 13.24, 8,240, 21.8, 330, 738, 121.8, 6, 5.28, 0.21, 30, 1.61, 0.15. We need to determine which of these values is relevant for the resistance and capacitance in the circuit.
Step 3: Calculating the Time Constant
The time constant \( \tau \) of an RC circuit is given by:
\[
\tau = R \times C
\]
where \( R \) is the resistance and \( C \) is the capacitance. The problem does not provide explicit values for \( R \) and \( C \), so we need to infer these from the given list.
Step 4: Determining the Lower Resistance Value
The problem asks for the "lower resistance value of the first capacitor," which is described as the "mind of the test class." This is likely a reference to the smallest value in the list, which could represent a resistance value.
Final Answer
The smallest value in the list is 0.15. Therefore, the lower resistance value of the first capacitor is:
\[
\boxed{0.15}
\]