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enter image description here

If the initial voltage of the capacitor at 0s is 20V, would the equation for the voltage of the capacitor at a specific time be; 60 + (20-60)e^(-t/RC)

or would it be 80 - 60e^(-t/RC)

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    \$\begingroup\$ It's unclear to me what the voltage is on the + terminal of your capacitor. It appears to be initially 0 volts given that - is at minus 20 and you appear to be stating that there is 20 volts across your capacitor. \$\endgroup\$
    – Andy aka
    Oct 31 at 9:34
  • \$\begingroup\$ The capacitor is initially charged to +20V- and the switch is closed at t=0s. In the other practice questions it is a full circuit. In this practice question it is not a full circuit and ends in -20V but is not explained why or what effect this has. \$\endgroup\$
    – james
    Oct 31 at 9:39
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    \$\begingroup\$ Then at t=0 the voltage on the left hand side of the capacitor is 0 volts. This is equivalent to charging the same capacitance (that is not pre-charged) and is connected to 0 volts on the RH side. \$\endgroup\$
    – Andy aka
    Oct 31 at 9:41
  • \$\begingroup\$ @Andyaka So would my initial formula of 60 + (20-60)e^(-t/RC) or would it be 80 - 60e^(-t/RC) \$\endgroup\$
    – james
    Oct 31 at 10:04
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Here are two scenarios: -

  • Capacitor fully discharged at t = 0 and lower terminal connected to 0 volts
  • Capacitor pre-charged to 20 volts with lower terminal connected to -20 volts: -

enter image description here

Notice that the exponential curve for both Vout1 and Vout2 are identical. At the left marker both Vout1 and Vout2 are 54.467 volts. At the right marker both are 59.953 volts. The curves are identical.

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