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schematic

simulate this circuit – Schematic created using CircuitLab

For a circuit like the above - why is the charge on the Capacitor right after the battery is added Capacitance * Voltage of the battery and not Capacitance * (Voltage of Battery - Voltage Drop due to Resistor)?

Isn't the potential difference on the Capacitor the Battery voltage - the voltage drop due to R1?

I'm doing book problems and it has a circuit set up this way and solutions state that the charge on the upper plate of this capacitor after contact with the battery is 1uF * 9V = 9uC.

Thanks for your time.

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  • \$\begingroup\$ For such a questions the only answer is "because of physics". It can be explained on different levels. Using analogies, math, models and so on. \$\endgroup\$ – Eugene Sh. Oct 20 '15 at 19:50
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There is only a voltage across the resistor when there is current flowing through it. Once the capacitor is charged up, then there's no current flowing.

When you first turn it on, there's no voltage on the capacitor, so there's 9V across the resistor, and hence 90mA flowing. This drops to nothing as the capacitor charges up.

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    \$\begingroup\$ Ahh, so the more the capacitor charges the less current there is and thus the voltage drop due to the resistor decreases as well? \$\endgroup\$ – Ulad Kasach Oct 20 '15 at 19:51
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    \$\begingroup\$ Yup. The product of the capacitor and the resistor values gives a 'time constant', which is a number that relates to the time taken to charge the capacitor. (Double the time constant = twice as long to charge) \$\endgroup\$ – user1844 Oct 20 '15 at 19:52
  • \$\begingroup\$ How long does the capacitor actually take to reach the 9V and charging current fall to zero? - multiple choice answer - (a) 1CR , (b) 5 CR, (c) 1000CR or (d) never. \$\endgroup\$ – JIm Dearden Oct 20 '15 at 20:13
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    \$\begingroup\$ b is good enough, less than .7% error. c is something like 10^-400. \$\endgroup\$ – Vladimir Cravero Oct 20 '15 at 20:46
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\$Q=CV_B\$ is only the steady state solution once the capacitor has become fully charged. When the battery is first connected, there is no charge on the capacitor and so the full potential of the battery falls across the resistor. This limits the current which flows as it begins to charge the capacitor. As the charge on the capacitor builds, the voltage across it begins to build. This means that the potential across the resistor, and therefore the charging current, decrease as the capacitor acquires more charge. The full equation for the charge on the capacitor at some time \$t\$ after the battery is connected (assuming it was initially uncharged) is: $$ Q=CV_B\left[1-e^{\frac{-t}{RC}} \right ] $$

As you can see, after a long time (a few time constants), the exponential term of the equation essentially vanishes and you are left with \$Q=CV_B\$.

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