1. **Problem statement:** A capacitor is charged to 12.0 V and connected to a voltmeter with internal resistance 3.40 M\Omega. After 4.00 s, the voltmeter reads 3.0 V. Find (a) the capacitance and (b) the time constant of the circuit. 2. **Relevant formula:** The voltage across a discharging capacitor is given by $$V(t) = V_0 e^{-t/\tau}$$ where $V_0$ is the initial voltage, $t$ is time, and $\tau = RC$ is the time constant. 3. **Step (a) Find the capacitance:** - Given: $V_0 = 12.0$, $V(4) = 3.0$, $t=4.00$ s, $R = 3.40 \times 10^6$ \Omega. - Use the formula: $$3.0 = 12.0 e^{-4/\tau} \Rightarrow \frac{3.0}{12.0} = e^{-4/\tau} = 0.25$$ - Taking natural log: $$\ln(0.25) = -\frac{4}{\tau} \Rightarrow -1.386 = -\frac{4}{\tau}$$ - Solve for $\tau$: $$\tau = \frac{4}{1.386} = 2.886 \text{ s}$$ - Since $\tau = RC$, solve for $C$: $$C = \frac{\tau}{R} = \frac{2.886}{3.40 \times 10^6} = 8.49 \times 10^{-7} \text{ F} = 0.849 \mu\text{F}$$ 4. **Step (b) Find the time constant:** - From above, $$\tau = 2.886 \text{ s}$$ **Final answers:** - (a) Capacitance $C = 0.849 \mu\text{F}$ - (b) Time constant $\tau = 2.89$ s (rounded)