Find the time constant for the given R C circuits in correct order. R_1=1 , R_2=2 , C_1=4 μ F, C_2=2 μ F

Correct answer: B. 18,8 / 9,4 aligned & =C R _1 & =(C_1+C_2)(R_1+R_2)=18 μ s _2 & =((C_1 C_2)/(C_1+C_2))((R_1 R_2)/(R_1+R_2))=(8)/(6) × (2)/(3)=(8)/(9) μ s _2 & =(C_1+C_2)((R_1 R_2)/(R_1+R_2))=(6)((2)/(3))=4 μ s aligned

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Physics
Chapter: Alternating Current

JEE Advanced 2006 · Physics · Alternating Current

Find the time constant for the given R C circuits in correct order....

\text{ Find the time constant for the given }

R C

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R_1=1 \Omega, R_2=2 \Omega, C_1=4 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}

Options

A $18,4,8 / 9$
B $18,8 / 9,4$
C $4,18,8 / 9$
D $4,8 / 9,18$

Correct answer

B. $18,8 / 9,4$

Step-by-step solution

\begin{aligned} \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} \end{aligned}

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