Question

Given, $R_1=1\Omega ,C_1=2\mu F$, $R_2=2\Omega ,C_2=4\mu F$ The time constants $($in $\mu S)$ for the circuits I, II, III are respectively

• A. 18, 8/9, 4
• B. 18, 4, 8/9
• C. 4, 8/9, 18
• D. 8/9, 18, 4

Answer 1

The correct option is D 8/9, 18, 4

For (I): The net capacitance $C=\frac{C_1{C}_2}{C_1+C_2}=\frac{2\times 4}{2+4}=4/3\mu F$ and net resistance $R=\frac{R_1{R}_2}{R_1+R_2}=\frac{1\times 2}{1+2}=2/3\Omega$

Thus, time constant $\tau =CR=\left(4/3\right)\left(2/3\right)=\left(8/9\right)\mu S$

For (II): $C=C_1+C_2=2+4=6\mu F$ and $R=R_1+R_2=1+2=3\Omega$

Thus, time constant $\tau =CR=\left(6\right)\left(3\right)=18\mu S$

For (III): $C=C_1+C_2=2+4=6\mu F$ and $R=\frac{R_1{R}_2}{R_1+R_2}=2/3\Omega$

Thus, time constant $\tau =CR=\left(6\right)\left(2/3\right)=4\mu S$