Question

An \(\mathrm{RC}\) circuit has a time constant of \(3.10 \mathrm{~s}\). At \(t=0,\) the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

Short answer

Based on the given information and analysis, the energy stored in the capacitor cannot reach exactly half of its maximum value. Therefore, it is not possible to find the time at which the energy stored in the capacitor reaches half of its maximum value in this case.

Step-by-step solution

Formula for the charge on a charging capacitor

First, we need to recall the formula for the charge on a capacitor as it charges: \(q(t) = Q_{max}(1 - e^{-\frac{t}{RC}})\) where \(q(t)\) is the charge at time \(t\), \(Q_{max}\) is the maximum charge the capacitor can hold, \(R\) is the resistance of the circuit, \(C\) is the capacitance, and \(t\) is time.

Formula for the energy stored in the capacitor

Now, we need to recall the formula for the energy stored in the capacitor: \(E(t) = \frac{1}{2}Cv^2(t)\) The voltage across the charging capacitor at a given time can be expressed in terms of the charge as follows: \(v(t) = \frac{q(t)}{C}\)

Substitute the charge for the voltage

Now we substitute the formula for the charge into the energy stored in the capacitor formula: \(E(t) = \frac{1}{2}C\left(\frac{q(t)}{C}\right)^2\) Simplifying this expression gives: \(E(t) = \frac{1}{2}\frac{q^2(t)}{C}\)

Express the energy in terms of maximum energy

We can now substitute the formula for charge from step 1 and express the energy stored in the capacitor at time \(t\) as: \(E(t) = \frac{1}{2}\frac{\left(Q_{max}(1 - e^{-\frac{t}{RC}})\right)^2}{C}\) The maximum energy stored in the capacitor can be expressed as: \(E_{max} = \frac{1}{2}\frac{Q_{max}^2}{C}\)

Set the energy to half of its maximum value and solve for t

Now we want to find the time when the energy stored in the capacitor reaches half of its maximum value, so we set \(E(t)\) to be half of \(E_{max}\): \(\frac{1}{2}E_{max} = \frac{1}{2}\frac{\left(Q_{max}(1 - e^{-\frac{t}{RC}})\right)^2}{C}\) Since we are interested in the time, we can divide both sides by \(\frac{1}{2}E_{max}\) to simplify the expression: \(1 = \frac{\left(Q_{max}(1 - e^{-\frac{t}{RC}})\right)^2}{Q_{max}^2}\) Taking the square root of both sides gives: \(\sqrt{1} = \frac{Q_{max}(1-e^{-\frac{t}{RC}})}{Q_{max}}\) Simplifying further, we get: \(1-e^{-\frac{t}{RC}}=\sqrt{1}\) Now, solve for the time \(t\): \(e^{-\frac{t}{RC}} = 1 - \sqrt{1}\) \(-\frac{t}{RC} = \ln(1 - \sqrt{1})\) \(t= -RC\ln(1 - \sqrt{1})\)

Substitute the given values

Finally, plug in the given time constant of the circuit, \(RC\) = 3.10 s: \(t = -3.10\ln(1-\sqrt{1})\) Since \(\sqrt{1}=1\), the expression inside the natural logarithm would be 0, and the natural logarithm of 0 is undefined. Therefore, the energy stored in the capacitor cannot reach exactly half of its maximum value in this case.