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I am trying to calculate the initial voltage across the capacitor. This textbook has used the Potential Divider formula to get the voltage across the 9ohm resistor (15V).

But I don't understand why the initial voltage across the capacitor is also 15V. Surely due to the 1ohm resistor in series with the source, there is a voltage drop across the 1ohm resistor, so there is less voltage stored in the capacitor.

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Consider the situation like this: the capacitor is an open circuit. That means that through the 1 ohm resistor, there is no current (which is also why we don't factor it in when calculating the voltage over the 9 ohm resistor). Since there is no current, there is also no voltage drop - V = I*R=0*1=0 V.

This is why the voltage over the capacitor is Vcap = 15-Vr = 15-0 = 15 V.

EDIT: As has been pointed out, the capacitor becomes an open circuit for DC only after some time, during which it charges. After that no current passes through the resistor as the capacitor is now fully charged.

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    \$\begingroup\$ Carefull the capacitor is only an open circuit to DC after some time. I prefer to state that the capacitor is fully charged with the result of no current in the resistor. \$\endgroup\$ – Decapod Nov 15 '16 at 22:05
  • \$\begingroup\$ @Decapod yes, that is a more accurate representation. \$\endgroup\$ – Egor Tamarin Nov 15 '16 at 22:17
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As it is a steady state, no current flows through the 1 Ohm resistor, when the capacitor is fully loaded. Therefore, no voltage drop can occur across the 1 Ohm resistor. The capacitor-voltage will converge to the 9/3 Ohm divider's voltage.

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In the given situation there are two conditions of importance.

1 The current trough the 3 ohm and 9 ohm resistor

  1. The knowledge that the capacitor has been charged and the current trough the 1 ohm resitor has stopped.

Since you understand the 15V over the 9 ohm it must be clear that the voltage over the capacitor is also 15V. Remember no current in the 1 ohm resistor and therefore no voltage drop.

Once you open the switch the capacitor starts to discharge over a resistor of 10 ohm ( 9 and 1 ohm in series).

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There will be a voltage drop across the 1 Ohm resistor while the capacitor is charging(t<0). However, When the capacitor has charged to the max value(t=0), it would behave as an open circuit. And because of this, there won't be any current flowing in the 1 Ohm resistor. The voltage across the 9 Ohm resistor would then be the voltage across Capacitor which is 15 Volts.

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