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I'm new to EE. I know this is a simple question, but I would like to know how the current flow in this circuit. My textbook says that the current does not pass through the capacitor, but I don't know why. Any help is appreciated, thanks!

schematic

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2 Answers 2

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enter image description here

Figure 1. A classic capacitor diagram. Image source: Electronics Tutorials.

A capacitor consists of two conductive plates, foils or conductive materials, separated by an insulator or dielectric. No current (other than some leakage) can pass from one plate to the other when run inside the designed parameters.

The symbol for the capacitor represents the two plates with a gap between them. DC current can't flow. Current does flow to charge or discharge the capacitor when the voltage across the plates changes. That's why capacitors can "pass" alternating current and their apparent impedance (AC resistance) decreases with higher frequency. (Incidentally, the resistor symbol represents a wirewound resistor and the inductor symbol represents a coil of wire.)

Tip: when analysing a circuit like that you can think of capacitors as blocking DC and passing high frequency. Inductors, on the other hand, will pass DC but block high frequency.

The linked article is worth reading.

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My textbook says that the current does not pass through the capacitor, but I don't know why.

Capacitors act as open circuits to a DC source (which is what you have), so no current can pass through C1. Since the capacitor doesn't pass current, R4 is not part of the circuit, and so current can only flow through R2->R3->R5. You can then calculate that current value using V1 and Ohm's law.

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  • \$\begingroup\$ When power is first applied, the capacitor will charge up to the voltage across R3. When the power is removed, the capacitor will discharge through R3 and R4. \$\endgroup\$ Jan 18, 2020 at 19:47
  • \$\begingroup\$ @PeterBennett True. But I think the answerer here was addressing only the OP's comment, "...my textbook says that the current does not pass through the capacitor..." That implies a steady-state perspective being demanded by the textbook. But your comment is also helpful in that the OP will soon need to step up to your point and understand it, too. \$\endgroup\$
    – jonk
    Jan 18, 2020 at 20:01
  • \$\begingroup\$ @jonk: I think I meant to put my comment under the original question - should pay more attention to where I post things! \$\endgroup\$ Jan 18, 2020 at 22:36
  • \$\begingroup\$ @Peter Bennett, In such circuits with reactive elements it should be clear if the voltage source is open or short circuit after "the power is removed". \$\endgroup\$ Jan 19, 2020 at 16:59

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