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enter image description here I am understanding a rc network shown in the above image . I need to calculate the Value Vc after i give 24 volts step input for 16 μs. i used capacitor charging equations Vc=Vinput(1-e^-t/rc) equation for outer loop but how do i take inner loop into calculation. Laplace method gives very huge polynomial . Could you please tell me if i can use time domain equation to calculate the value of Vc ? i am stuck on this problem for last 2 days . Kindly Help me out folks .

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    \$\begingroup\$ It's not clear from your question that you realise you can't consider the outer loop in isolation, i.e. it's not simply a 24V step and a time constant, \$\small \tau=261k\Omega\: \times 470pF\$. This is because the inner 'loop' affects \$\small V_C\$ \$\endgroup\$ – Chu Jan 24 at 15:12
  • \$\begingroup\$ exactly @Chu . so there will be some effect by inner loop. How do i write its equation ? Can you give any hints about it \$\endgroup\$ – Rohan Jan 24 at 15:14
  • \$\begingroup\$ You've tried Laplace - that's one method. The other method is to derive the differential equations and solve simultaneously. The amount of effort is about the same by either method. So it's whichever method you're most comfortable with. \$\endgroup\$ – Chu Jan 24 at 15:19
  • \$\begingroup\$ ok. but how do i write differential equation for inner loop ? can you just give an example ? @Chu \$\endgroup\$ – Rohan Jan 24 at 15:27
  • \$\begingroup\$ Assume \$\small I_1\$ and \$\small I_2\$ flow through \$\small C_1\$ and \$\small C_2\$,respectively. Then use KCL/KVL to form two simultaneous equations, then solve. I don't have the time! \$\endgroup\$ – Chu Jan 24 at 15:35
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\$ F(s) = \frac{V_C}{V_1} = k \cdot \frac{1+a_1 \cdot s}{1+b_1 \cdot s+b_2 \cdot s^2}\$

\$k = \frac{R_2+R_3}{R_1+R_2+R_3}\$

\$a_1 = \frac{R_2 R_3 C_1}{R_2+R_3}\$

\$b_1 = \frac{C_1 R_3 (R_1+R_2) +C_2 R_1 (R_2+R_3)}{R_1+R_2+R_3}\$

\$b_2 = \frac{C_1 C_2 R_1 R_2 R_3}{R_1+R_2+R_3}\$

Inverse Laplace of the step response (\$F(s) \cdot \frac{24}{s}\$) at \$t=16 \cdot 10^{-6}\$ gives 2.26538 V.

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    \$\begingroup\$ While this answer is useful for finding the answer, it could be more useful if you detailed to the user how you reached this finding. \$\endgroup\$ – loudnoises Jan 25 at 9:03
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    \$\begingroup\$ I start with impedance Z1(s) which is defined as 1/(sC1) parallel R3 plus series resistor R2. Next impedance is Z2(s), which is Z1(s) parallel 1/(sC2). Resulting voltage is Vc=V1*Z2(s)/(Z2(s)+R1). \$\endgroup\$ – UweD Jan 25 at 9:10
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    \$\begingroup\$ Basically following this approach: en.wikipedia.org/wiki/Voltage_divider \$\endgroup\$ – UweD Jan 25 at 9:16
  • \$\begingroup\$ @UweD I appreciate your Big help and I elaborated the given steps on paper . \$\endgroup\$ – Rohan Jan 26 at 18:57

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