Question

In the given circuit , $R_1=1\Omega ;R_2=2\Omega ;C_1=4\mu \text{F};C_2=2\mu \text{F}$, then the time constant (in $\mu \text{s}$) for the given $\text{RC}$ circuits are

• A. $4,\frac{8}{9},18$
• B. $4,18,\frac{8}{9}$
• C. $18,\frac{8}{9},4$
• D. $18,4,\frac{8}{9}$

Answer 1

The correct option is C $18,\frac{8}{9},4$

Time constant , $\tau =CR$

For $\text{CASE-I}:$

$C=C_1+C_2=4+2=6\mu \text{F}$ and $R=R_1+R_2=1+2=3\Omega$

$\therefore {\tau }_1=6\times 3=18\mu \text{s}$

For $\text{CASE-II}:$

$C=\frac{C_1{C}_2}{C_1+C_2}=\frac{4\times 2}{4+2}=\frac{8}{6}=\frac{4}{3}\mu \text{F}$ and $R=\frac{R_1{R}_2}{R_1+R_2}=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega$

$\therefore {\tau }_2=\frac{4}{3}\times \frac{2}{3}=\frac{8}{9}\mu \text{s}$

For $\text{CASE-III}:$

$C=C_1+C_2=4+2=6\mu \text{F}$

$R=\frac{R_1{R}_2}{R_1+R_2}=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega$

${\tau }_3=6\times \frac{2}{3}=4\mu \text{s}$

Hence, option (c) is the correct answer.