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i always knew that current will take all paths, even the other post on this forum said so, but i was doing this RC circuit today where the resistor and the capacitor are wired in parallel, i was told that the capacitor charges quickly because the current " will the take the path of least resistance", so now i am confused

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    \$\begingroup\$ How about the path of least impedance? A capacitor has complex impedance that is a function of frequency. \$\endgroup\$ – George Herold Feb 17 '15 at 21:15
  • \$\begingroup\$ i am pretty sure she said the path of least resistance, moreover, i was just taking this course, the instructor also said something about the current taking the shortest path to the ground \$\endgroup\$ – user28324 Feb 17 '15 at 21:17
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    \$\begingroup\$ Just read it as "will mostly take the path of least resistance". \$\endgroup\$ – The Photon Feb 17 '15 at 21:17
  • \$\begingroup\$ i didnt understand yet :D so it might take but not always? \$\endgroup\$ – user28324 Feb 17 '15 at 21:21
  • \$\begingroup\$ Have you learned about frequency-domain analysis yet, or are you trying to understand time-domain transient analysis? \$\endgroup\$ – MarkU Feb 17 '15 at 21:23
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The current will mostly flow through the capacitor at the beginning because at time=0 the capacitor isn't full and basically looks like a short to the circuit. Since it looks like a short to the circuit, it charges very quickly. As the capacitor charges, it starts to look like an open circuit though. You could think of it as a variable resistor that starts at 0 and goes up to infinite as you charge it.

Once the capacitor is charged, the resistance is infinite and all of the current will flow through the resistor instead of the capacitor.

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The behavior of a capacitor is governed by the equation: \begin{align} I = C \frac{dV}{dt} \end{align}

Think about what happens in the limit as \$\frac{dV}{dt} \rightarrow \infty\$: I must also tend to infinity. This is equivalent to saying that the capacitor "shorts out" any rapid changes in voltage.

When you initially hook a capacitor to a voltage source, you are providing a step voltage, of which the slope at the step is "infinite" (technically it's undefined, but there's no such thing as a perfect step in the real world), so the capacitor initially acts as a short.

So what would actually happen in the real world? Well, in the real world things have resistance (either resistance in the wires connecting the source, internal to the source, or even internal to the capacitor). This limits the maximum current that can flow into the capacitor, which means that the rate of change of the voltage across the capacitor is also limited (i.e. the voltage across the capacitor can only change so fast).

So how does this behavior translate into a more complicated circuit (your parallel RC circuit)? One way to analyze this is to look at the "complex impedance" of the capacitor. \begin{align} Z_C = \frac{1}{j\omega C} \end{align} Where \$\omega\$ is the frequency of the voltage across the capacitor (in rad/s) and j is the imaginary number \$\sqrt{-1}\$. Let's ignore this j for now and only consider the magnitude of \$Z_C\$: \begin{align} |Z_C| = \frac{1}{\omega C} \end{align} If you work out the units, you'll realize that \$Z_C\$ has units of Ohms, the same as resistance! This is not an accident, and means we can apply the familiar Ohm's law with the RC circuit and work out how much current flows through the capacitor and how much flows through the resistor by comparing the relative magnitudes of the resistor's resistance and the magnitude of the capacitor's impedance (for a certain frequency).

A step input is composed of many frequencies, tending towards infinity (search for "Fourier series" to learn more), so there are some components which have an 'infinite frequency'. What does this do for \$Z_C\$? Well, it drives \$Z_C\$ for those high frequencies towards 0, i.e. a short (equivalent to the phenomenological hand-waving I did first).

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  • \$\begingroup\$ This is way beyond the comprehension of OP obviously, but it is a good addition to what horta says for those that want to actually know why stuff happens. +1 \$\endgroup\$ – I. Wolfe Feb 17 '15 at 22:14
  • \$\begingroup\$ your comment is way out of my league :D but i really appreciate the effort, thanks \$\endgroup\$ – user28324 Feb 17 '15 at 22:48
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Imagine your power supply is a motor rotating at constant speed but disconnected mechanically from a large static flywheel. At some point you activate the coupling mechanism and instantly the motor stalls as all its rotational energy is drained by the static flywheel. For a short period of time the motor is stalled but gradually it starts ramping up the speed of the flywheel until some time later the motor and flywheel are rotating at the motor's original off-load speed.

A discharged. capacitor behaves the same when connected across a dc supply. Instantly the output voltage drops to zero as all the energy is sapped by the cap but, after a period of time the cap becomes charged and attains the original voltage of the power supply.

The resistor is incidental.

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Current does take all possible paths, but not necessarily equally. In your RC circuit, current will flow through both elements, but significantly more may flow through the capacitor as it charges up.

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  • \$\begingroup\$ yea, but why? the prof said it's treated like the resistor is not their \$\endgroup\$ – user28324 Feb 17 '15 at 21:30
  • \$\begingroup\$ Current is always determined by the voltage difference, in this case the supply voltage (whatever specified) vs. the capacitor's voltage (which is zero when discharged). Putting resistors across the capacitor doesn't change the voltage--unless there's resistance in series, and without a schematic, I have to assume there is none. \$\endgroup\$ – gbarry Feb 17 '15 at 23:00
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My grandfather used to refer to electricity as "juice". "Turn on the 'juice'!" I thought of this as I was trying to visualize an analogy to demonstrate the parallel RC circuit: An elevated cistern full of water has the potential energy (battery). A large pipe (low resistance wire) comes down from the cistern with the flow controlled by a valve (switch). The large pipe empties into a closed barrel in parallel with a smaller pipe (resistance). When the barrel is empty, the flow from the cistern is great (high current), the barrel fills fast, and there is little or no pressure to feed the smaller pipe (resistance). As the barrel fills, the flow slows ever more gradually, and the flow increases in the smaller pipe. Eventually, the flow to the barrel (capacitor) stops, and the only flow is the small, constant flow through the smaller pipe. This one-way flow is like direct current. The time it takes for the barrel to reach a certain percent of "full" (capacity or pressure (voltage)), is it's "time constant". The time constant is determined by the size of the barrel and the plumbing. I hope this analogy is not too flawed and I'd be interested in the opinion of the learned people as to if this holds water.

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  • \$\begingroup\$ Is "holds water" a pun? This is pretty accurate to how the circuit actually works. \$\endgroup\$ – Greg d'Eon Feb 27 '15 at 14:42

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