91影视

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.

TheRCtime constant is evaluated using the formula:

$蟿=RC$

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

The two possible resistance are,

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

The two possible capacitance is calculated as,

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

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

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

Therefore, the required values are:

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