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## Hot answers tagged electric-circuits

7

If you have just given the voltage signal with $$\def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V$$ then the current at $t=2\rm s$ is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that  i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ...

5

You are absolutely correct, the electric field does fall off with distance from the battery. However, this is only true during the transient state (the state of the field when the battery is first connected). In fact not only are the magnitudes inconsistent, but so is the direction of the field. The field doesn't always point in the direction of the wire. ...

3

The capacitor will indeed be charged a little -- but the charge will be so low that we may as well call it uncharged. Here is why: the open switch is another capacitor (two conducting terminals, although not quite in plate form, separated by a dielectric). Its capacitance is extremely low, though: the terminals' cross section will be on the order of a ...

2

In figure (a), the voltage is continuous but the time derivative is not; the capacitor current would discontinuously change sign from positive to negative. In figure (b) however, the voltage is discontinuous. It is typically said that the voltage across an ideal capacitor is continuous since, for the current to exist, the time derivative of the voltage ...

2

In Nilsson and Riedels textbook: Electric circuits, it is actually stated on page 28 that " when represented in a circuit diagram, copper or aluminum wiring isn't usually modeled as a resistor; the resistance of the wire is so small compared to the resistance of the other elements in the circuit that we can neglect the wiring resistance to simplify the ...

1

There's no reason the sides have to be equal, but if they aren't, the capacitor obviously has a net electric charge. Moreover, the electric field lines emanating from the capacitor have to go somewhere, such that the whole capacitor is also one half of a larger capacitor. In a circuit model, you would simply represent this as two or more separate capacitors, ...

1

You are right in principle, but by considering the field of a battery; you are considering something, which is considered negligible by your book. In real world problems, one always try to ignore effects which are negligible, to solve problems to a reasonable level of accuracy and highlight the principles. If your book starts to talk of all these effects, ...

1

Increasing the resistance of a circuit does not necessarily increase the reading on your voltmeter. It depends on whether you use a voltage source or a current source. E.g.: You circuit is powered by a 1.5V battery you will read 1.5 Volts when measuring on the ends of the circuit even if you increase the resistance. But if you measure the current it will ...

1

What might help (or not) here is to distinguish between reference polarity and absolute polarity. When you label one terminal of the inductor with a plus sign, you're simply choosing a reference polarity. You are, in essence, choosing which end to place the "red lead" of your voltmeter. If you calculate that the voltage across the inductor is negative ...

1

Well, if you look at a practical circuit which produces that graph, there must be a sudden drastic change in the circuit at that instant of time to cause the capacitor to abruptly shift mode from charging to discharging - quite possibly a switch/switches were turned on/off effectively putting the capacitor in a different circuit. If you are asking the ...

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