Q90 / 120 · JEE Advanced 2006 · Physics
$$
\text { Find the time constant for the given } R C \text { circuits in correct order. }
$$
$R_1=1 \Omega, R_2=2 \Omega, C_1=4 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}$
$R_1=1 \Omega, R_2=2 \Omega, C_1=4 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}$
- A$18,4,8 / 9$
- Correct answerB$18,8 / 9,4$
- C$4,18,8 / 9$
- D$4,8 / 9,18$
Solution
$$
\begin{aligned}
\tau & =C R \\
\tau_1 & =\left(C_1+C_2\right)\left(R_1+R_2\right)=18 \mu \mathrm{s} \\
\tau_2 & =\left(\frac{C_1 C_2}{C_1+C_2}\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=\frac{8}{6} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s} \\
\tau_2 & =\left(C_1+C_2\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=(6)\left(\frac{2}{3}\right)=4 \mu \mathrm{s}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.