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enter image description here

For the above circuit, I calculate that the equivalent resistance of the charging circuit is 2.668 kΩ, and the equivalent capacitance is 2.055 µF. Therefore, the time constant would be 2.668*2.055 = 5.48 ms as the amount of time required to charge. I simulated the circuit and got the following time constant:

enter image description here

It is almost twice as doubled, while it should be approximately the same. Can anyone explain to me the difference? Thank you.

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    \$\begingroup\$ Did you forget R4? \$\endgroup\$ – Brian Drummond Apr 8 '14 at 4:57
  • \$\begingroup\$ @BrianDrummond I thought R4 was for when the circuit is discharging. \$\endgroup\$ – Vu Chau Apr 8 '14 at 5:05
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    \$\begingroup\$ You appear to have R4 in the wrong place. I think you intended it be in the circuit only when the switch is in the down position. Instead, it's in series when switch is in the up position, and produces a confusing result. \$\endgroup\$ – gwideman Apr 8 '14 at 5:20
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As drawn R4 and R3 are in series making the equivalent charging resistor

\$ R = \frac{1.62 \cdot 1.78}{1.62 + 1.78} + 1.82 + 2.43 = 5.098 k\Omega\$

and capacitor

\$ C = \frac{1.4 \cdot 2.2}{1.4 + 2.2} + 1.2 = 2.055 \mu F\$

The time constant is therefore

\$\tau = C \cdot R = 10.47 \text{ ms}\$

As pointed out in comments you have neglected R4. If R4 is only supposed to effect the discharge time it should be connected between the switch and earth.

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