Question

In the figure shown below, $R_1=1\Omega ,R_2=2\Omega ,C_1=4\mu \text{F},C_2=2\mu \text{F}$ , the time constants (in $\mu \text{s}$) for the circuits $I,II$ and $III$ respectively are

• A. $18,4,\frac{8}{9}$
• B. $18,\frac{8}{9},2$
• C. $18,\frac{8}{9},4$
• D. $4,\frac{8}{9},18$

Answer 1

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

All the three circuits given are $\text{RC}$ circuits. Therefore, the time constant for each circuit is $RC$.

Finding the equivalent resistance and capacitance for all three given circuits,

For first circuit:

$R_{eq}^{\prime }=1+2=3\Omega ;C_{eq}^{\prime }=4+2=6\mu \text{F}$

For second circuit:

$R_{eq}^{\prime \prime }=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega ;C_{eq}^{\prime \prime }=\frac{4\times 2}{4+2}=\frac{4}{3}\mu \text{F}$

For third circuit:

$R_{eq}^{\prime \prime \prime }=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega ;C_{eq}^{\prime \prime \prime }=4+2=6\mu \text{F}$

Time constant in first circuit , ${\tau }_1=R_{eq}^{\prime }{C}_{eq}^{\prime }=18\mu \text{s}$

Time constant in second circuit , ${\tau }_2=R_{eq}^{\prime \prime }{C}_{eq}^{\prime \prime }=\frac{8}{9}\mu \text{s}$

Time constant in third circuit , ${\tau }_3=R_{eq}^{\prime \prime \prime }{C}_{eq}^{\prime \prime \prime }=4\mu \text{s}$

Hence, option (c) is the correct answer.