Let's solve the problem step-by-step as requested. --- ## **Given:** - Battery Voltage, \( V = 9\,\text{V} \) - Resistor, \( R = 3\,\Omega \) - Capacitor, \( C = 10000\,\mu\text{F} = 10000 \times 10^{-6}\,\text{F} = .01\,\text{F} \) --- ## **1. The time constant of the RC circuit** The time constant (\( \tau \)) of an RC circuit is the product of resistance and capacitance. --- ## **2. Formula for time constant** \[ \tau = R \cdot C \] --- ## **3. Substitution of values** \[ \tau = 3\,\Omega \times .01\,\text{F} = .03\,\text{seconds} \] --- ## **4. Expression for capacitor voltage during charging** The voltage across the capacitor as it charges is: \[ V_c(t) = V \left(1 - e^{-t/\tau}\right) \] Where: - \( V \) = supply voltage - \( t \) = time - \( \tau \) = time constant --- ## **5. Calculate the capacitor voltage after \( t = 3\,\text{s} \)** Plug in the values: \[ V_c(3) = 9 \left(1 - e^{-3/.03}\right) \] Calculate \( -\frac{3}{.03} = -100 \): \[ V_c(3) = 9 \left(1 - e^{-100}\right) \] Since \( e^{-100} \) is extremely close to zero (practically zero): \[ V_c(3) \approx 9 \times (1 - ) = 9\,\text{V} \] --- ## **Summary Table** | Step | Calculation/Expression | Value/Result | |------|----------------------------------------------------|-------------------| | 1 | Time constant formula | \( \tau = R \cdot C \) | | 2 | Substitute values | \( \tau = 3 \times .01 \) | | 3 | Time constant value | \( \tau = .03\,\text{s} \) | | 4 | Capacitor charging voltage | \( V_c(t) = V(1 - e^{-t/\tau}) \) | | 5 | At \( t = 3\,\text{s} \): \( V_c(3) \) | \( \approx 9\,\text{V} \) | --- ### **Final Answer** - **Time constant:** \( .03\,\text{s} \) - **Capacitor voltage at \( t = 3\,\text{s} \):** \( 9\,\text{V} \) (to three significant figures) **Explanation:** After 3 seconds (which is 100 times the time constant), the capacitor is essentially fully charged and reaches the supply voltage.