Let's determine the time constant (\(\tau\)) for the given RC circuit. ### **Step 1: Identify the Equivalent Resistance Seen by the Capacitor** The time constant \(\tau\) is given by: \[ \tau = R_{eq} \cdot C \] #### **Analyzing the Circuit:** - **R4 and R5** are in series with the capacitor \(C\). - **R2** is in parallel with the series combination of **R4, R5,** and **C**. - **R1** and **R3** are in series with the above network. #### **Let's Simplify:** 1. **R4 and R5 in series:** \[ R_{45} = R_4 + R_5 = 6\ \Omega + 6\ \Omega = 12\ \Omega \] 2. **R2 in parallel with R45:** \[ R_{245} = \frac{R_2 \cdot R_{45}}{R_2 + R_{45}} = \frac{80 \cdot 12}{80 + 12} = \frac{960}{92} = 10.43\ \Omega \] 3. **R1 and R3 in series with R245:** \[ R_{total} = R_1 + R_3 + R_{245} = 60 + 60 + 10.43 = 130.43\ \Omega \] #### **But the time constant only depends on the resistance "seen" by the capacitor.** When the capacitor is charging, the resistance in the path directly in series with the capacitor must be considered. The resistors **R4** and **R5** are directly in series with the capacitor. However, this combination is in parallel with **R2**. So, the resistance seen by the capacitor is the equivalent resistance of **R2** in parallel with (**R4** + **R5**), and this combination is in series with **R1** and **R3**. \[ R_{eq} = R_1 + R_3 + \left( \frac{R_2 \cdot (R_4 + R_5)}{R_2 + R_4 + R_5} \right) \] Plug in the values: \[ R_1 = 60\ \Omega, \quad R_2 = 80\ \Omega, \quad R_3 = 60\ \Omega, \quad R_4 = 6\ \Omega, \quad R_5 = 6\ \Omega \] \[ R_{eq} = 60 + 60 + \left( \frac{80 \cdot (6+6)}{80 + 6 + 6} \right) \] \[ R_{eq} = 120 + \left( \frac{80 \cdot 12}{92} \right) \] \[ R_{eq} = 120 + \left( \frac{960}{92} \right) \] \[ R_{eq} = 120 + 10.43 = 130.43\ \Omega \] ### **Step 2: Time Constant** \[ \tau = R_{eq} \cdot C = 130.43\ \Omega \cdot C \] --- ### **Final Answer:** \[ \boxed{ \tau = 130.43\ \Omega \cdot C } \] where \(C\) is the capacitance value. If you have the value of \(C\), you can substitute it in.