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: Identify the resistive network.
The circuit consists of two resistors of value $R$ connected in parallel.
The equivalent resistance of two equal resistors in parallel is
\[ R_{\text{eq}} = \frac{R \cdot R}{R + R} = \frac{R}{2} \]
Step 2: Identify the capacitive network.
The circuit also consists of two capacitors of value $C$ connected in parallel.
The equivalent capacitance of two capacitors in parallel is
\[ C_{\text{eq}} = C + C = 2C \]
Step 3: Write the expression for time constant.
The time constant of an RC network is given by
\[ \tau = R_{\text{eq}} \times C_{\text{eq}} \]
Step 4: Substitute the equivalent values.
\[ \tau = \left(\frac{R}{2}\right) \times (2C) \]
\[ \tau = RC \]
Step 5: Conclusion.
Hence, the time constant of the given network is
\[ \boxed{CR} \]

Answer 2

Since a resistor-capacitor network's time constant is fixed by \(\tau = R_{eq}C_{eq}\), and here two equal resistors \(R\) appear in parallel while two equal capacitors \(C\) appear in parallel, an alternative route is to first reduce the resistors together and the capacitors together, then check the resulting product against each candidate answer.

Two resistors of value \(R\) in parallel give \(R_{eq} = R/2\), and two capacitors of value \(C\) in parallel give \(C_{eq}=2C\). Multiplying these equivalents together: \[ \tau = R_{eq}\,C_{eq} = \frac{R}{2}\times 2C = RC \]

1. \(CR\): This matches the product obtained from combining the parallel resistors and parallel capacitors separately.
2. \(2CR\): This would result only if the capacitors combined to \(4C\) instead of \(2C\), which is not the case for two equal capacitors in parallel.
3. \(CR/4\): This would result only if the resistors combined to \(R/4\), which would require four equal resistors in parallel rather than two.
4. \(CR/2\): This would result if either the resistor combination or the capacitor combination were halved beyond what parallel combination of two equal elements actually gives.

Combining the resistor pair and the capacitor pair by the standard parallel formulas, and multiplying the two equivalents, gives exactly \(RC\), with the factor of 2 from halving the resistance cancelling the factor of 2 from doubling the capacitance.

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