JEE Advanced · Physics · 22. AC Circuits
Find the time constant for the given R C circuits in correct order.

\(R_1=1 \Omega, R_2=2 \Omega, C_1=4 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}\)
- A \(18,4,8 / 9\)
- B \(18,8 / 9,4\)
- C \(4,18,8 / 9\)
- D \(4,8 / 9,18\)
Answer & Solution
Correct Answer
(B) \(18,8 / 9,4\)
Step-by-step Solution
Detailed explanation
\(\tau =C R \)
\( \tau_1 =\left(C_1+C_2\right)\left(R_1+R_2\right)=18 \mu \mathrm{s} \)
\( \tau_2 =\left(\frac{C_1 C_2}{C_1+C_2}\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=\frac{8}{6} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s} \)
\( \tau_2 =\left(C_1+C_2\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=(6)\left(\frac{2}{3}\right)=4 \mu \mathrm{s}\)
\( \tau_1 =\left(C_1+C_2\right)\left(R_1+R_2\right)=18 \mu \mathrm{s} \)
\( \tau_2 =\left(\frac{C_1 C_2}{C_1+C_2}\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=\frac{8}{6} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s} \)
\( \tau_2 =\left(C_1+C_2\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=(6)\left(\frac{2}{3}\right)=4 \mu \mathrm{s}\)
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