Question

How long will it take, as a multiple of the time constant, \(\tau,\) for the capacitor in an RC circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)

Short answer

Answer: d) \(4 \tau\)

Step-by-step solution

Figure out the equation for the capacitor charging process in an RC circuit

The charging process of a capacitor in an RC circuit can be given by the formula: $$ Q(t) = Q_{max}(1 - e^{-\frac{t}{\tau}}) $$ where \(Q(t)\) is the charge on the capacitor at time \(t\), \(Q_{max}\) is the maximum possible charge on the capacitor, \(\tau\) is the time constant, and \(t\) is the time.

Set up the equation for 98% charge

Since we want to find the time it takes for the capacitor to be 98% charged, we will use the equation to set up the proportion: $$ Q(t) = 0.98Q_{max} $$

Solve the equation for the time t

Substitute the charging process equation for the 98% charge equation: $$ Q_{max}(1 - e^{-\frac{t}{\tau}}) = 0.98Q_{max} $$ We can now solve this equation for the time \(t\). Divide both sides by \(Q_{max}\): $$ 1 - e^{-\frac{t}{\tau}} = 0.98 $$ Rearrange the equation to get the exponential term on its own: $$ e^{-\frac{t}{\tau}} = 0.02 $$ Now, take the natural logarithm of both sides: $$ -\frac{t}{\tau} = ln(0.02) $$ Rearrange the equation to find \(t\): $$ t = -\tau ln(0.02) $$ Calculate t: $$ t \approx 3.912 \tau $$

Identify the closest answer from the given options

From the given options, we can see that the closest answer to our calculated value of \(t\) is: d) \(4 \tau\) Thus, it will take approximately \(4 \tau\) for the capacitor in an RC circuit to be 98% charged.