Two figures are shown as Fig. A and Fig. B. The time constant of Fig. A is $\tau_{\mathrm{A}}$ and time…

Two figures are shown as Fig. A and Fig. B. The time constant of Fig. A is $\tau_{\mathrm{A}}$ and time constant of Fig. Bis $\tau_{\mathrm{B}}$. Then
  1. $\tau_{\mathrm{A}}=\frac{1}{4} \mathrm{~s}$ and $\tau_{\mathrm{B}}=5 \mathrm{~s}$
  2. $\tau_{\mathrm{A}}=\frac{1}{2} \mathrm{~s}$ and $\tau_{\mathrm{B}}=\frac{1}{5} \mathrm{~s}$
  3. $\tau_{\mathrm{A}}=4 \mathrm{~s}$ and $\tau_{\mathrm{B}}=5 \mathrm{~s}$
  4. $\tau_{\mathrm{A}}=4 \mathrm{~s}$ and $\tau_{\mathrm{B}}=5 \mathrm{~s}$

Solution

For circuit A , $\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=4+\frac{6 \times 12}{6+12}=8 \Omega, \mathrm{~L}_{\mathrm{eq}}=2 \mathrm{H} \\ & \therefore \quad \tau_{\mathrm{A}}=\frac{\mathrm{L}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}}=\frac{2}{8}=\frac{1}{4} \mathrm{~s} \end{aligned}$
For circuit B, $\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{10 \times 10}{10+10}=5 \Omega, \mathrm{C}_{\mathrm{eq}}=0.5+0.5=1 \mathrm{~F} \\ & \therefore \quad \tau_{\mathrm{B}}=\mathrm{R}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}=5 \times 1=5 \mathrm{~s} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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