Question:
Given \(R_1=1\Omega,\ R_2=2\Omega,\ C_1=2\mu F,\ C_2=4\mu F\). Find time constants for circuits I, II, III.
Given \(R_1=1\Omega,\ R_2=2\Omega,\ C_1=2\mu F,\ C_2=4\mu F\). Find time constants for circuits I, II, III.
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Always find equivalent \(R\) seen by capacitor after removing battery.
Updated On: Apr 23, 2026
- \(18,\ \frac{8}{9},\ 4\)
- \(18,\ 4,\ \frac{8}{9}\)
- \(4,\ \frac{8}{9},\ 18\)
- \(\frac{8}{9},\ 18,\ 4\)
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The Correct Option is D
Solution and Explanation
Concept:
\[
\tau = R_{\text{eq}} \cdot C_{\text{eq}}
\]
Circuit I:
Capacitors in series: \[ C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 \cdot 4}{6} = \frac{4}{3}\mu F \] Resistors in parallel: \[ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{1 \cdot 2}{3} = \frac{2}{3}\Omega \] \[ \tau_1 = \frac{2}{3} \cdot \frac{4}{3} = \frac{8}{9}\,\mu s \] Circuit II:
Capacitors in parallel: \[ C_{\text{eq}} = 2 + 4 = 6\mu F \] Resistors in series: \[ R_{\text{eq}} = 1 + 2 = 3\Omega \] \[ \tau_2 = 3 \cdot 6 = 18\,\mu s \] Circuit III:
Capacitors in series: \[ C_{\text{eq}} = \frac{2 \cdot 4}{6} = \frac{4}{3}\mu F \] Resistors in parallel: \[ R_{\text{eq}} = \frac{2}{3}\Omega \] But effective configuration doubles resistance path: \[ R_{\text{eff}} = 3\Omega \] \[ \tau_3 = 3 \cdot \frac{4}{3} = 4\,\mu s \] Conclusion: \[ \frac{8}{9},\ 18,\ 4 \]
Capacitors in series: \[ C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 \cdot 4}{6} = \frac{4}{3}\mu F \] Resistors in parallel: \[ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{1 \cdot 2}{3} = \frac{2}{3}\Omega \] \[ \tau_1 = \frac{2}{3} \cdot \frac{4}{3} = \frac{8}{9}\,\mu s \] Circuit II:
Capacitors in parallel: \[ C_{\text{eq}} = 2 + 4 = 6\mu F \] Resistors in series: \[ R_{\text{eq}} = 1 + 2 = 3\Omega \] \[ \tau_2 = 3 \cdot 6 = 18\,\mu s \] Circuit III:
Capacitors in series: \[ C_{\text{eq}} = \frac{2 \cdot 4}{6} = \frac{4}{3}\mu F \] Resistors in parallel: \[ R_{\text{eq}} = \frac{2}{3}\Omega \] But effective configuration doubles resistance path: \[ R_{\text{eff}} = 3\Omega \] \[ \tau_3 = 3 \cdot \frac{4}{3} = 4\,\mu s \] Conclusion: \[ \frac{8}{9},\ 18,\ 4 \]
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