Question

The time constant of the network shown in the figure is 

• A. $CR$
• B. $2CR$
• C. $CR/4$
• D. $CR/2$

Hint:

For time constant calculations, always first reduce the circuit to its equivalent resistance and equivalent capacitance as seen by the source.

Answer 1

The Correct Option is A

Step 1: Combine the two parallel resistors: \(R_{eq} = R \parallel R = R/2\).
Step 2: Combine the two parallel capacitors: \(C_{eq} = C + C = 2C\).
Step 3: Multiply to get the time constant: \(\tau = R_{eq}C_{eq} = (R/2)(2C)\).
\[ \boxed{\tau = CR} \]

Answer 2

A quick scaling argument also confirms this result. Replacing a single resistor \(R\) with two identical resistors in parallel halves the effective resistance, while replacing a single capacitor \(C\) with two identical capacitors in parallel doubles the effective capacitance. Since the time constant of a simple RC network is the product of its resistance and capacitance, halving one factor while doubling the other leaves the product unchanged from what a single \(R\) and a single \(C\) alone would give.

1. \(CR\): This is exactly what a single \(R\)-single \(C\) network would give, and the halving/doubling from using two of each cancels out, reproducing this same value.
2. \(2CR\): This would require the two scaling factors to reinforce each other instead of cancelling, which does not happen here since one factor halves and the other doubles.
3. \(CR/4\): This would require both factors to shrink the product, but doubling the capacitance works in the opposite direction to shrinking.
4. \(CR/2\): This would require only one of the two scalings to apply, whereas both the resistor-halving and capacitor-doubling occur together and cancel each other.

Therefore, the correct answer is \(CR\).