Question

A $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{}\mathbf{and}\mathbf{}\mathbf{a}\mathbf{}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{-}\mathbf{\mu F}$capacitor can be connected in series or parallel, as can a $\mathbf{25}\mathbf{.}\mathbf{0}$- and a $\mathbf{100}\mathbf{-}\mathbf{k\Omega }$ resistor. Calculate the four RCtime constants possible from connecting the resulting capacitance and resistance in series.

Short answer

The values are obtained as:

$$\begin{gathered} {\mathrm{\tau }}_1=0.1975\mathrm{s} \\ {\mathrm{\tau }}_2=1.1875\mathrm{s}\mathrm{s} \\ {\mathrm{\tau }}_3=0.0316\mathrm{s} \\ {\mathrm{\tau }}_4=0.199\mathrm{s} \end{gathered}$$

Step-by-step solution

Define circuit

A circuit is a closed-loop through which electrons can pass. Electrical energy is provided in the circuit by a source of electricity, such as a battery. No electrons will travel unless the circuit is complete, that is, it returns to the electrical source in a full circle.

Evaluating the RC time constants

TheRCtime constant is evaluated using the formula:

$\tau =RC$

Where the value of R is resistance in the circuit and the value of C is capacitance.

The two possible resistance are,

localid="1656393507362" $$\begin{gathered} R_s=R_1+R_2 \\ =\left(2.50\times {10}^4\Omega +1.00\times {10}^5\Omega \right) \\ =1.25\times {10}^5\Omega \\ {R}_p=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{\left(2.50\times {10}^4\Omega \right)\times \left(1.00\times {10}^5\Omega \right)}{\left(2.50\times {10}^4\Omega \right)+\left(1.00\times {10}^5\Omega \right)} \\ =2.00\times {10}^4\Omega \end{gathered}$$

The two possible capacitance is calculated as,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{s}}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2} \\ =\frac{\left(2.00\times {10}^{-6}\mathrm{F}\right)\times \left(7.50\times {10}^{-6}\mathrm{F}\right)}{\left(2.00\times {10}^{-6}\mathrm{F}\right)+\left(7.50\times {10}^{-6}\mathrm{F}\right)} \\ =1.58\times {10}^{-6}\mathrm{F} \\ {\mathrm{C}}_{\mathrm{p}}={\mathrm{C}}_1+{\mathrm{C}}_2 \\ =\left(2.00\times {10}^{-6}\mathrm{F}\right)+\left(7.50\times {10}^{-6}\mathrm{F}\right) \\ =9.5\times {10}^{-6}\mathrm{F} \end{gathered}$$

Finally, the four-time constant can be evaluated as,

$$\begin{gathered} {\tau }_1=R_s{C}_s \\ =\left(1.25\times {10}^5\Omega \right)\times \left(1.58\times {10}^{-6}F\right) \\ =0.197s \\ {\tau }_2=R_s{C}_p \\ =\left(1.25\times {10}^5\Omega \right)\times \left(9.5\times {10}^{-6}F\right) \\ =1.1875s \\ {\tau }_3=R_p{C}_s \\ =\left(2.00\times {10}^4\Omega \right)\times \left(1.58\times {10}^{-6}F\right) \\ =0.0316s \\ {\tau }_4=R_p{C}_p \\ =\left(2.00\times {10}^4\Omega \right)\times \left(9.5\times {10}^{-6}F\right) \\ =0.19s \end{gathered}$$

Therefore, the required values are:

$$\begin{gathered} {\tau }_1=0.1975s \\ {\tau }_2=1.185ss \\ {\tau }_3=0.0316s \\ \tau 4=0.19s \end{gathered}$$