Question

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=0,\) the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega,\) that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q ?\)

Short answer

Question: Determine the time for the capacitor to charge to half of its maximum value, the ratio of energy stored in the capacitor to its maximum possible value at this instant, the time constant for discharging with a new resistor, and the time for the capacitor to discharge half of its maximum stored charge for a series RC circuit with resistance R = 10.0 Ω, capacitance C = 10.0 μF, and maximum voltage V_max. Answer: The time for the capacitor to charge to half of its maximum value is approximately 6.93 x 10^-5 seconds. The ratio of energy stored in the capacitor to its maximum possible value at this instant is 0.25. The time constant for discharging with a new resistor is 10^-5 seconds, and the time for the capacitor to discharge half of its maximum stored charge is approximately 6.93 x 10^-6 seconds.

Step-by-step solution

Determine the time constant for charging

The time constant for charging a capacitor in an RC circuit is given by \(\tau = RC\). Given the values \(R = 10.0 \Omega\) and \(C = 10.0 \mu F\), we can calculate the time constant \(\tau\): $$\tau = RC = (10.0 \Omega)(10.0 \times 10^{-6} F) = 10^{-4} s$$

Find the time for the capacitor to be charged to half of its maximum value

The voltage across the capacitor at a given time is given by \(V(t) = V_{max}(1 - e^{(-t/\tau)})\). We need to find the time \(t\) for which the capacitor is charged to half of its maximum value, i.e. \(V(t) = \frac{V_{max}}{2}\). We can rewrite the equation as: $$e^{-t/\tau} = 1 - \frac{V_{max}}{2V_{max}}$$ Now, we can find the time \(t\) in terms of the time constant \(\tau\): $$t = -\tau \ln(1 - \frac{1}{2}) = \tau \ln(2)$$ Using the value of \(\tau\) from Step 1, the time it takes to charge the capacitor to half of its maximum value is: $$t = 10^{-4} \ln(2) \approx 6.93 \times 10^{-5} \, s$$

Find the ratio of energy stored in the capacitor to its maximum value

At this instant, the energy stored in the capacitor is given by \(U(t) = \frac{1}{2}CV^2\), whereas the maximum energy stored is \(U_{max} = \frac{1}{2}CV_{max}^2\). The ratio of these energies is: $$\frac{U(t)}{U_{max}} = \frac{\frac{1}{2}CV^2}{\frac{1}{2}CV_{max}^2} = \left(\frac{1}{2}V_{max}\right)^2$$ So the ratio of the energy stored in the capacitor to its maximum possible value is \(1/4\) or \(0.25\).

Calculate the time constant for discharging with a new resistor

For discharging, the time constant is given by \(\tau' = R'C\). The new resistor is \(R' = 1.00 \Omega\). So the time constant for discharging is: $$\tau' = R'C = (1.00 \Omega)(10.0 \times 10^{-6} F) = 10^{-5} s$$

Find the time for the capacitor to discharge half of its maximum stored charge

The charging equation can be adjusted for discharging by changing the sign of the exponent: \(Q(t) = Q_{max}(e^{-t/\tau'})\). To find the time it takes for the capacitor to discharge half of its maximum stored charge, we can write: $$\frac{1}{2}Q_{max} = Q_{max}e^{-t/\tau'}$$ Dividing by \(Q_{max}\) and solving for \(t\), we get: $$t = -\tau' \ln\left(\frac{1}{2}\right) = \tau' \ln(2)$$ Using the value of \(\tau'\) from Step 4, the time it takes for the capacitor to discharge half of its maximum stored charge is: $$t = 10^{-5} \ln(2) \approx 6.93 \times 10^{-6} \, s$$