Question

The voltage $V(t)$ across a capacitor of capacitance $C$ discharging through a resistance $R$ is given by $V(t) = V_0e^{-t/RC}$. The initial voltage $V_0$ is 10.00 $\pm$ 0.02 V, $R = 100.0 \pm 0.1 \ M\Omega$ and $C = 1.00 \ \mu F$. The circuit is closed and subsequently disconnected after 100 $\pm$ 1 seconds. The final voltage is best represented as

• A. 2.71 $\pm$ 0.01 V
• B. 3.67 $\pm$ 0.003 V
• C. 3.67 $\pm$ 0.05 V
• D. 3.67 $\pm$ 0.37 V

Answer 1

Answer: (C)

3.67 $\pm$ 0.05 V

Answer 2

The voltage across a discharging capacitor is given by $V(t) = V_0e^{-t/RC}$.

Given values: Initial voltage $V_0 = 10.00 \pm 0.02$ V Resistance $R = 100.0 \pm 0.1 \ M\Omega = (100.0 \pm 0.1) \times 10^6 \ \Omega$ Capacitance $C = 1.00 \ \mu F = 1.00 \times 10^{-6} \ F$ (Uncertainty in C is not given, so assume it's negligible) Time $t = 100 \pm 1$ s

1. Calculate the nominal value of the final voltage $V(t)$:

First, calculate the time constant $\tau = RC$. $\tau = (100.0 \times 10^6 \ \Omega) \times (1.00 \times 10^{-6} \ F) = 100.0 \ s$

Now, substitute the values into the voltage equation: $V(t) = V_0e^{-t/RC} = 10.00 \times e^{-100/100.0} = 10.00 \times e^{-1}$ $V(t) = 10.00 \times 0.36787944... \approx 3.67879 \ V$ Rounding to two decimal places, $V(t) \approx 3.68 \ V$. Looking at the options, $3.67 \ V$ is preferred.

2. Calculate the uncertainty in the final voltage $\Delta V$:

The formula for error propagation for a function $Y = f(X_1, X_2, \dots, X_n)$ is given by: $(\Delta Y)^2 = \left(\frac{\partial Y}{\partial X_1} \Delta X_1\right)^2 + \left(\frac{\partial Y}{\partial X_2} \Delta X_2\right)^2 + \dots + \left(\frac{\partial Y}{\partial X_n} \Delta X_n\right)^2$

In our case, $V = V_0e^{-t/RC}$. We need to find $\Delta V$ considering uncertainties in $V_0$, $t$, and $R$. We assume $\Delta C = 0$.

The partial derivatives are: $\frac{\partial V}{\partial V_0} = e^{-t/RC}$ $\frac{\partial V}{\partial t} = V_0 e^{-t/RC} \left(-\frac{1}{RC}\right) = -\frac{V}{RC}$ $\frac{\partial V}{\partial R} = V_0 e^{-t/RC} \left(-\frac{t}{C}\right) (-R^{-2}) = V \frac{t}{CR^2}$

Now, substitute these into the error propagation formula: $(\Delta V)^2 = \left(\frac{\partial V}{\partial V_0} \Delta V_0\right)^2 + \left(\frac{\partial V}{\partial t} \Delta t\right)^2 + \left(\frac{\partial V}{\partial R} \Delta R\right)^2$ $(\Delta V)^2 = \left(e^{-t/RC} \Delta V_0\right)^2 + \left(-\frac{V}{RC} \Delta t\right)^2 + \left(V \frac{t}{CR^2} \Delta R\right)^2$

Divide by $V^2$ to get the relative uncertainty squared: $\left(\frac{\Delta V}{V}\right)^2 = \left(\frac{e^{-t/RC} \Delta V_0}{V}\right)^2 + \left(\frac{-\frac{V}{RC} \Delta t}{V}\right)^2 + \left(\frac{V \frac{t}{CR^2} \Delta R}{V}\right)^2$ Since $V = V_0e^{-t/RC}$, the first term simplifies: $\frac{e^{-t/RC}}{V} = \frac{e^{-t/RC}}{V_0e^{-t/RC}} = \frac{1}{V_0}$. The last term simplifies: $\frac{t}{CR^2} \Delta R = \frac{t}{RC} \frac{\Delta R}{R}$.

So, $\left(\frac{\Delta V}{V}\right)^2 = \left(\frac{\Delta V_0}{V_0}\right)^2 + \left(\frac{\Delta t}{RC}\right)^2 + \left(\frac{t}{RC} \frac{\Delta R}{R}\right)^2$

Now, calculate the relative uncertainties: $\frac{\Delta V_0}{V_0} = \frac{0.02}{10.00} = 0.002$ $\frac{\Delta t}{t} = \frac{1}{100} = 0.01$ $\frac{\Delta R}{R} = \frac{0.1}{100.0} = 0.001$

And the terms in the equation: $t/RC = 100/100 = 1$ $\frac{\Delta t}{RC} = \frac{\Delta t}{t} \times \frac{t}{RC} = 0.01 \times 1 = 0.01$ $\frac{t}{RC} \frac{\Delta R}{R} = 1 \times 0.001 = 0.001$

Substitute these values into the equation for $\left(\frac{\Delta V}{V}\right)^2$: $\left(\frac{\Delta V}{V}\right)^2 = (0.002)^2 + (0.01)^2 + (0.001)^2$ $\left(\frac{\Delta V}{V}\right)^2 = 0.000004 + 0.000100 + 0.000001$ $\left(\frac{\Delta V}{V}\right)^2 = 0.000105$

Now, calculate $\frac{\Delta V}{V}$: $\frac{\Delta V}{V} = \sqrt{0.000105} \approx 0.01024695$

Finally, calculate the absolute uncertainty $\Delta V$: $\Delta V = V \times \left(\frac{\Delta V}{V}\right)$ Using $V \approx 3.67879 \ V$: $\Delta V = 3.67879 \times 0.01024695 \approx 0.03769 \ V$

Rounding $\Delta V$ to one or two significant figures, it is approximately $0.04 \ V$ or $0.038 \ V$. The nominal voltage is $3.67879 \ V$. If we round it to $3.67 \ V$, then the uncertainty should be consistent. Comparing our result with the options: (A) 2.71 $\pm$ 0.01 V (B) 3.67 $\pm$ 0.003 V (C) 3.67 $\pm$ 0.05 V (D) 3.67 $\pm$ 0.37 V

Our calculated voltage is $3.67 \ V$ (or $3.68 \ V$). Our calculated uncertainty is $0.03769 \ V$. Option (C) has $V = 3.67 \ V$ and $\Delta V = 0.05 \ V$. This is the closest option to our calculated uncertainty. $0.03769 \ V$ rounded up to one significant figure for uncertainty would be $0.04 \ V$. If we round to a multiple of 5 for simplicity, $0.05 \ V$ is a reasonable choice if the precision is lower.

Let's re-evaluate the rounding. Usually, the uncertainty is quoted to one or two significant figures. $0.03769 \approx 0.038$ or $0.04$. The options are $0.003, 0.05, 0.37$. $0.05$ is the closest option to $0.038$ or $0.04$.

The final voltage is best represented as $3.67 \pm 0.05 \ V$.