I am having a hard time with capacitors and the role they play in circuits. Lets say a circuit has a capacitor with C = 3 microF and a resistor of 3 ohms and a battery with a voltage of 12 V. How would you find the current in the circuit? How does the current flowing through the circuit be affected by the charging of the capacitor? What happens to the current when the capacitor is fully charged? .
Thanks!
Figuring out the current in an RC circuit is likely beyond the scope of the MCAT. For this explanation, I will assume you are talking about a direct current (DC) circuit with the resistor and capacitor
in series. Basically, you always have to refer to Kirchoff's loop rule, which states that if you make a loop in a circuit, the potential drops/gains must sum to zero:
V - IR - Q/C = 0
(remember that the voltage drop across a capacitor is equal to Q/C)
So consider the point when you first connect the battery to the circuit. The charge Q on the capactior is intially zero, so Q/C = 0 and therefore V - IR = 0. So the initial current can simply be calculated by using V = IR. In your example, the current I would be 4 amps.
As the current continues to flow, charge will accumulate on the plates. This accumulating charge will repel any other incoming charges. As a result, the current decreases. This can be seen from the loop rule, since as Q increases, Q/C increases. Since V and R are constant, I must decrease!
The current will continue to decrease while the current is flowing. In fact, the current after a very long time will approach zero (though it will never actually reach zero, just become really, really close).
Hopefully this explanation makes sense. If you need a further explanation please let me know!
