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Proportionality in RC circuits

Physics Asked by PeaceSeeker127eq on January 13, 2021

I know that the time constant $tau= RC$, where $R$ is the resistance and $C$ is the capacitance.
But I’m confused, bear with me, when I double or increase the capacitance of the RC circuit.

If I double the capacitance of the RC circuit, will the time constant be also double, or half of its original value? Can the same be said if I also do it to the resistor?

Thank you for taking the time.

2 Answers

If you double the capacitance, then your new capacitance is $$ C_{new} = 2C, $$ whereas the new time constant is $$ tau_{new} = RC_{new}=2RC. $$

Answered by Vadim on January 13, 2021

As already stated, from the math, if you increase the capacitance or resistance you increase the time constant RC. However, it is equally important to understand what the time constant is and why increasing R or C increases the time constant, and vice versa.

The time constant is how long it takes to charge the capacitor, through the resistor, from an initial charge voltage of zero to approximately 63.2% ($1-e^{-1}$) of an applied DC voltage, or to discharge the capacitor through the same resistor to 37.8% ($e^{-1}$) of its initial charge voltage.

The relationship between capacitance, voltage, and charge is

$$C=frac{Q}{V}$$

So the greater the capacitance the greater the amount of charge it will have on its plates for a given voltage. That, in turn, means for a larger capacitor it takes longer to add or remove the charge in order to achieve the same increase or decrease in voltage, all else being equal.

For a given capacitance and voltage the greater the resistance the lower the current. Since current is the charge per unit time added to or removed from the capacitor, a lower current increases the time it takes to charge or discharge the capacitor, I.e., increases the time constant. Likewise a lower resistance and higher current decreases time constant.

Hope this helps.

Answered by Bob D on January 13, 2021

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