Q177 / 200 · Manipal (MET) 2014 · Physics

Given, $\begin{array}{ll} R_1=1 \Omega & C_1=2 \mu \mathrm{~F} \\ R_2=2 \Omega & C_2=4 \mu \mathrm{~F} \end{array}$ Figure
The time constant (in $\mu \mathrm{s}$ ) for the circuits I, II, III are respectively

  • A
    $18,8 / 9,4$
  • B
    $18,4,8 / 9$
  • C
    $4,8 / 9,18$
  • Correct answer
    D
    $8 / 9,18,4$

Solution

For (I): The net capacitance \(\mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{2 \times 4}{2+4}=4 / 3 \mu \mathrm{~F}\) and net resistance \(\mathrm{R}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=\frac{1 \times 2}{1+2}=2 / 3 \Omega\) Thus, time constant \(\tau=\mathrm{CR}=(4 / 3)(2 / 3)=(8 / 9) \mu \mathrm{S}\) For (II): \(\mathrm{C}=\mathrm{C}_1+\mathrm{C}_2=2+4=6 \mu \mathrm{~F}\) and \(\mathrm{R}=\mathrm{R}_1+\mathrm{R}_2=1+2=3 \Omega\) Thus, time constant \(\tau=\mathrm{CR}=(6)(3)=18 \mu \mathrm{~S}\) For (III): \(\mathrm{C}=\mathrm{C}_1+\mathrm{C}_2=2+4=6 \mu \mathrm{~F}\) and \(\mathrm{R}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=2 / 3 \Omega\) Thus, time constant \(\tau=\mathrm{CR}=(6)(2 / 3)=4 \mu \mathrm{~S}\)

Hence, the correct answer is option: 'D'.

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