Question

A condenser of some internal resistance $r$ , when connected in series with a resistance $R_1$, the time constant is found to be ${\tau }_1$. When it is connected in series with a resistance $R_2$, the time constant is found to be ${\tau }_2$. The capacitance is:

• A. $\frac{{\tau }_1-{\tau }_2}{R_1-R_2}$
• B. $\frac{{\tau }_1{R}_1({\tau }_2-{\tau }_1)}{{\tau }_2}$
• C. $\frac{{\tau }_2{\tau }_1\left(R_2\right)}{({\tau }_2-{\tau }_1)}$
• D. $\frac{{({\tau }_2-{\tau }_1)}^2{R}_1^2}{{\tau }_2(R_1-R_2)}$

Answer 1

The correct option is A $\frac{{\tau }_1-{\tau }_2}{R_1-R_2}$

$R_1$ connected in series then it's effective resistance will be $r+R_1$

${\tau }_1=C(r+R_1)$

$\frac{{\tau }_1}{C}=r+R_1$ ---(1)

$R_2$connected in series then it's effective resistance will be $r+R_2$

${\tau }_2=C(r+R_2)$

$\frac{{\tau }_2}{C}=r+R_2$ ---(2)
from 1 & 2 ,

$C=\frac{{\tau }_1-{\tau }_2}{R_1-R_2}$