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It is a analogic useful way to understand the resistive circuit, in the meaning that the electric expression $V = RI$ has the same form as the mechanical $F = kv$, in an environment where the drag force is proportional to the velocity. In the case of a conductor, it is important to note that even whithout any applied field, the free electrons have momentum ...


4

What you wrote as Maxwell's equations are valid for free charges $\rho$, free currents $J=\rho v$, in vacuum and in these there is no Ohm's law involved. If instead of free charges/currents in vacuum you have macroscopic bulk material, fluid, gas, etc., your version of Maxwell's equations do not describe those. Instead we assume that we have some knowledge ...


4

Exactly as @Claudio Saspinski says. I was about to write an answer in those same terms when I saw his excellent answer. Think of the metal as an extraordinarily dense "fluid" in which particles moving (electrons) almost immediately acquire their limit velocity. The faster they want to go, the more resistance they find, in proportion to their own ...


3

Here is an answer to Chittaranjan rout's second question. Yes indeed, you will form a capacitor exactly as you have described it, and at the instant you connect the wires to a cell, a very tiny burst of current will flow out along the wires and charge them up very quickly. Because the size of the wires is small, the capacitance of a parallel pair of wires ...


3

First a zener diode is nothing but a diode, just a heavily doped one at that (Let's get to that later). When you apply a voltage across a diode, you get an I-V characteristic like this (Source:Google, Electronics notes) What we need to focus on is the reverse bias part. Suppose you apply a reverse biasing voltage. An electron crossing the depletion region (...


3

a potential difference created by a battery causes charge carriers to flow because of an electric force, which can be represented by having an electric field everywhere in the circuit. A potential difference created by a battery only causes charge carriers to flow if there is a load connected to the battery terminals. If there is no load connected, there is ...


3

First, notice that the supplied current of the current source, must be equal to the current that flows through the 'back' of it, and therefore must be equal to the current that flows 'upward' from node $C$: Then, applying KCL on node $C$: Yields: $$I_1 = I_1 +I_2$$ $$I_2=0A$$


3

If I understand what you are saying correctly, $\overrightarrow{J}$ is a known quantity and so you only have $6$ scalar quantities to worry about - not $9$. This is all slightly complicated by the fact that the equations don't just include $\overrightarrow B$ and $\overrightarrow E$, they also contain their time derivatives - $\frac{\partial \overrightarrow{ ...


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Hint: This following circuit can be thought of as Wheatstone bridge, in which the middlemost resistor has no current flowing akin to your $2R$, in the question.


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You can. 1/total resistance = 1/resistance 1 + 1/resistance 2 You can work out the total resistance, and therefore the total current. and you can work out the power from there. You need to understand the basics of parallel circuits


2

As @Bahrudin Trbalic already pointed out, if there were zero resistance the initial current to an uncharged ideal capacitor when switched on to a voltage source would be theoretically be infinite. Fact is, however, there is always resistance in the circuit (with the exception of a superconductor). As a practical matter, in designing equipment one always ...


2

There will be a transfer of electrons to the empty capacitor. If there is no resistance, the current will be infinite which is not good (and not possible). $$ I = \frac{dq}{dt} = C\frac{dV}{dt} $$ also $$ I = \frac{V}{R} $$ so if $R = 0 $, you can see the problem.


1

To see this, let's cut the wires from $B$ to $G$ and from $D$ to $G$ for a moment. Notice that $B$ is exactly halfway between $A$ and $C$ (in terms of resistance along that path), $D$ is also halfway, and so is $G$. That means they are all at exactly half the potential difference between $A$ and $C$. Connecting these three points with a wire will not have ...


1

AC does imply that the current is changing direction (usually in a periodic fashion). DC does not have to be constant. It can change with time (as in the slow charging of a capacitor).


1

You may derive the phase difference in mixed circuits using complex numbers. You CAN add Vr and Vl to get Esource, but all values have to be at the same instance. You can't simply add their amplitudes as they are out phase. So you have to use vector addition(phasor is based on concept of vectors) to calculate the source voltage.


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Perhaps this example will be useful: In the world of radio engineering, it is important to understand how much radio frequency energy being put out by a transmitter can be transformed into electromagnetic waves by an antenna connected to that transmitter. For best power transfer, the antenna is sized so it resonates as a half-wave at the frequency at which ...


1

Assuming it's not deliberate, two things come to mind: light-sensitive switches will have significant hysteresis to avoid flickering on and off in dim light. So they will require brighter light to switch on in the morning than to switch off in the evvening. human vision perceives a given light level as being brighter during a transition from dark to ...


1

He mentions that the shape of the cross section is irrelevant but it should be the same across the resistor. So why [is this assumption] necessary? Also is it necessary that cross sections be perpendicular to the z-axis? These assumptions are necessary to simplify the problem so that the field lines are straight (so that a simple relationship between ...


1

In my answer to your previous question of the same title, I described how current can flow around and not through the dielectric of a capacitor. To avoid confusion, you should change the title of this question because it is a duplicate of the title of your previous related question. The capacitor only looks like an open circuit if the voltage across it is ...


1

If you have a cell, a bulb and a capacitor connected in series then this is not an open circuit (do not be misled by the circuit diagram symbol for a capacitor). Current will flow from the cell to charge the capacitor, and as the current passes through the bulb, it will light the bulb (as long as there is enough current to make the bulb filament glow). ...


1

The capacitor is an open circuit for dc. As such it is sometimes called a blocking capacitor in equivalent circuits where you need to decouple the dc bias voltages from the ac signals..


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This one does look tricky at first glance. However, you can use the symmetry of the circuit to come to a conclusion. Because the outer paths (going through the top and right; and through the bottom and left resistors, respectively), are completely equivalent, the current through these two paths must be the same. But then the voltage drop across the top $R$ ...


1

Your approach is correct, for the most part. Consider the leftmost circuit. Since, as you said, $R_2$ is short-circuited, we can redraw the leftmost circuit: Thus, applying Kirchhoff's Current Law on the blue node yields $i_1=0$. No current flows through $R_2$, since it is short circuited. The voltage across the CCVS is indeed $2i_1=2\cdot0=0V$, so yes, ...


1

When the wiper is turned as you describe, the wiper is essentially touching the top lug on the potentiometer so the resistance between them is zero. This means that the voltage drop between the top lug and the wiper is also zero, which means the wiper lug is at the same voltage as the top lug. Then, turning the pot down from there inserts resistance between ...


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