A $70\mu F$ capacitor is discharged through two parallel resistors,$15K\Omega$and$25K\Omega$ By what factor will the time constant of this circuit increase if the resistors are instead placed in series with each other?

We need to find what factor will the time constant of this circuit increase if resistors are instead placed in series with each other.

when the switch is off, the capacitors discharge and the current flows through the two resistors the time taken to discharge the capacitor is called the time constant $\tau$and it is given by equation $\left(28.30\right)$in the form

$\tau =RC$

Since R is resistance, and C is capacitance in the circuit when the resistors are connected one by one and with no junction between them. They are called to be connected in series for a series resistor, the equivalent resistance for their combination is given by equation $\left(28.15\right)$in the form

$$\begin{gathered} R_{series}=R_1+R_2 \\ =15k\Omega +25k\Omega \\ =40k\Omega \end{gathered}$$

If there are two points and there are resistors connected at both ends of the points, we called that the resistors are connected in parallel. Equation localid="1649266247543" $\left(28.24\right)$shows the equivalent resistance for the parallel connection in the form

$$\begin{gathered} \frac{1}{R_{parallel}}=\frac{1}{15K\Omega }+\frac{1}{25k\Omega } \\ {R}_{parallel}={\left(\frac{1}{15k\Omega }+\frac{1}{25k\Omega }\right)}^{-1} \\ {R}_{parallel}=9.375k\Omega \end{gathered}$$

As shown by equation$\left(1\right)$ the time constant is directly proportional to resistance. So, to get the ratio $\left({\tau }_{series/}{\tau }_{parallel}\right)$by

$\frac{{\tau }_{scries}}{{\tau }_{parallel}}=\frac{R_{series}}{R_{parallel}}$

Let us Plug the values for $R_{parallel}$and $R_{series}$into equation $\left(2\right)$to get ratio,

role="math" localid="1649266708835" $$\begin{gathered} \frac{{\tau }_{series}}{{\tau }_{parallel}}=\frac{R_{series}}{R_{parallel}} \\ =\frac{40k\Omega }{9.375k\Omega } \\ =4.3. \end{gathered}$$