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-1

in the absence of the voltage V2, R2 and R3 will be parallel so R(equivalent)=(R3*R2)/(R3+R2) ,and R(equivalent) which im gonna call R' is in serie with R1 so you circuit will contain just the voltage V1 and a resistor R''=R'+R1. in the presence of the voltage V2, R1,R2 and R3 are in wye "Y" ,get a look to this ...


1

The resolution is that currents can change instantaneously. Remember that the reason that the current in a single solenoid cannot change instantaneously is that a change in current causes a change in flux, which causes an electromotive force (emf) to oppose the original change in current. However, suppose now there are two currents, as in your example. ...


1

note: I accidentally thought OP was asking about a series $LC$, not a series $LCR$. Including the $R$ changes the results here by making the infinities turn into large finite values. Suppose you hook your series $LC$ circuit up to a voltage source with frequency dependent phasor $\tilde{V}_s(\omega)$. Intuition First let's guess what happens. At low ...


0

Say your series RLC circuit is excited by a constant current source producing current $I$ at angular frequency $\omega$. Since all 4 elements (R, L, C, and source) are in series, the current through any one of them is just the source current. Then for each element, $V_n=IZ_n$. For an inductor, the impedance Z is given by $Z=i\omega{}L$. For a capacitor, ...


0

Please look here you can make a function of w for both voltage across inductor and capacitor, and you can check the neighbourhood of resonant freq.


0

For a series RL circuit with DC source and switch, there is a problem with opening the switch after it has been closed for some time. In the context of ideal circuit theory, the current through an inductor must be continuous since the voltage across is proportional the time derivative of the current through. Put less rigorously, if the inductor current is ...


1

First of all, I'll write the energy with the letter $U$ to not confuse with the electric field $E$. By definition of power, $P=\frac{dU}{dt}$. Now, where is this energy in the equation coming from? $$U=\int \vec{F}\cdot \vec{dl} = q \int \vec{E}\cdot \vec{dl} = qV$$ This is the amount of energy gained by the charge when moving across an electric potential ...


1

First equation: power is defined as energy per unit time. Second equation: if a current flows through a circuit, the power dissipated is the product of voltage and current. This is because the voltage describes the energy each electron is given to traverse the circuit, and the current describes the number of electrons that travel the circuit per unit time. ...


0

As noted by Floris, the best way to measure small-value resistance is to use a four-point Kelvin connection; unless the current drawn by the voltage meter is significant (in which case there are other problems) then provided that current-source probes are either the inner two or the outer two (as opposed to being interleaved with the voltage-reading ones) ...


2

Ok so I first have taken the diagram from the wikipedia page for reference and put it here. Now if you are happy with the idea of how the potential divider works... .... then I hope that you can see that $R_1$ and $R_2$ in the Wheatstone Bridge diagram form a potential divider and there is another potential divider with $R_3$ and $R_x$ - and the points ...


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Short answer - yes, everything in the circuit can contribute. But usually, an ohmmeter is zeroed with the probes in place - in other words, whatever resistance the probes represent is taken out by the meter. There are two other factors that play a role, especially when you try to measure small resistance. The first of these is contact resistance: it is ...


2

Since the lamps are in series, the electric current through each is identical; all of the current out of one lamp is in to the other lamp; if there is a flow through one, there is a flow through the other. It cannot be that there is a flow through one and not the other (in the context of this simple model). Thus, if the lamps are identical, their ...


2

The current is the conventional current in the opposite direction to the electrons current but they are the same thing , the current is the same in series connections. If the two lamps are identical they will give the same amount of light.


1

In response to your comment, there is a way to find the voltage without finding equivalent resistance. You can write the potentials at each junction point. Taking the points on the lower line to be zero potential, the potential gain on going up the last branch is $1$V. Then going left, add another $1$V. Then, current through the last branch is 1A and the ...


3

should the voltage be the input voltage or the voltage across that specific resistor The latter. generally when you want to know something about a particular component, you work with the conditions applying at the boundary of that specific component. I have a desk lamp with a LED. The other end of the lamp plugs into a 240 V AC power outlet. If I want ...


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The amount of current that goes on either branch is dependent upon the resistance and any voltage potentials of the branch. Kirchhoff's law states that the same amount of current that enters a node, must leave the node.


0

Why must it contain a resistor The author is most likely referring to the radiation resistance. A circuit delivers electrical energy to a resistor where it is converted to heat, i.e., the energy is not stored but is lost to the environment. Analogously, a circuit delivers electrical energy to an antenna where it is converted to electromagnetic ...


0

The capacitor is connected in parallel with the bleeder resistor. The voltage between the terminals of the capacitor is equal to the potential difference between the bleeder resistor. So the volatage should......?


-1

Hint: The 10 micro farad capacitor can be removed ignored since it gets stuck in a wheatstone like network.


1

You will have to use Kirchoff's law to get the answer. How can some of positive numbers be zero? No, the sum of all charges will be zero while that of positive plate will be finite. Now the net sum would be zero since charge is conserved on the system having all the right plates. Use Kirchoff's Loop Law and Kirchoff's current law to find charges and ...


0

A conductor is made of neutral atoms, which has equal numbers of protons and electrons. The protons and some of the electrons are fixed. But some of the electrons, the "valence" electrons, can move freely in the conductor. Eletrons repel each other, so they tend to spread out as much as they can without leaving the conductor. If you add extra electrons, ...


1

EMF is the total voltage that can be supplied by a source of electrical energy (e.g. battery/dynamo). Basically, it encompasses both the voltage that will reflect in the circuit and the voltage that is constantly used up to overcome the resistance r of the component itself. Voltage is equivalent to work done per charge (V= W/q), this applies to EMF as work ...


0

The emf of the battery causes a potential drop across the circuit due to the chemical reactions inside the battery. When a charge gets to the positive end of the battery(remember that in reality,in most cases it is the electrons that flow in a circuit and are moving from the negative end to the positive end of the battery) they get attracted by that positive ...


0

In DC, the split ring commutator is what allows the current in the loop (i.e. the bit inside the magnetic field) to change direction every half turn, so that the force acting on the loop always acts to turn it in the same direction. The current is direct in DC regardless of the split ring commutator. The AC motor has alternating current, but I'm not sure ...


1

There is no need for taking the mod of the charge. The voltage has the same sign as the charge. So if you start out with +20 µC on one capacitor (they give the + sign for a reason - so that's the side where we will put the positive charge) and +60 µC on the other (from $Q=CV$) then it follows that the total of redistributed charge is 80 µC as you correctly ...


2

My guess; you are mixing up quadripoles and quadrupoles. Quadripoles are two-port networks used in electric circuit analysis. The original German word is "Vierpol Theorie", which means Four-pol because of 4 Poles. https://en.wikipedia.org/wiki/Two-port_network Quadrupoles are related to multipole expansion used in electromagnetic, atomic orbital,.. theory. ...


1

Trying to address this misconception: I start of with a resistance of 1 ohm by the wire and 6 amperes which result in 6 volts. When I meet the resistor however the resistance increases to let's say 3. Does the current decrease at the same rate the resistance increases? So if the resistance goes down to 3 will the current be 2 so that in the end I have a ...


0

There's an energy transfer whenever there is a change in potential, not potential difference. The (electric) potential, measured in volts, is the electric potential energy (EPE) of a unit charge at a particular point in the circuit. So, imagine a particle of unit charge travelling in the direction of current. It starts off with a higher potential (therefore ...


1

But isn't this solution wrong? I think it should be wrong, because we are talking about AC here; That is, 220V is peak voltage, not mean voltage. I think that mean voltage is something below 220V, thus resistance is going to be below 484 Ohms. If you do not live in central or north america, 220V will be RMS, so your teacher is right.


0

As you already mentioned $$ I = \int \vec{J} \cdot d\vec{A} $$ current is the charge-flow through a given surface. So If one talks about currents, the surface (and therefore its normal direction) is to be understood beforehand. You could assign a direction like this: $$ \frac{\vec{I}}{I} = \frac{\int d\vec{A}}{A} $$ which is the mean normal direction of the ...


0

What I think is the best way to look at it is that current is completely local. What I mean is that actually current is defined a distinct region of space. We care about how much charge there is in a region of space and then how much charge is in that same region without worrying to what happened to the first amount of charge. On the other hand, the current ...


1

A few points of confirmation/correction: Yes the electrons flow in a direction opposite to the "conventional" current. No there is no "deficiency" if electrons - rather they have a different "potential" which is caused by the chemical reactions in the battery. No you don't have to invoke "surface charges" in the wire in order to understand current - ...


0

Consider the case where we just have a battery and a resistor in series and nothing else. Lets say the battery gives the electron some amount of energy $U$. Now look at one electron as it travels around the circuit. Let's say the battery gives the electron some amount of energy $U$. If the electron still had some energy left when it arrived back at the ...


2

Kirchoff's voltage law. Quoting Wikipedia, The directed sum of the electrical potential differences (voltage) around any closed network is zero, or: More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop, or: The algebraic sum of the products of the resistances of the conductors ...


1

So what exactly happens to the potential inside the resistor ? Unlike the ideal conductors, for which an electric field cannot exist inside, there is an electric field through the resistor body when there is a current through. And, as you may know, the rate of change in electric potential is related to the value of the electric field. Thus, the ...


3

Let me first take a little detour away from this circuit to particle accelerators. If you have some electrons in vacuum and a potential set up between two points (exactly the same as saying you have an electric field set up) you can accelerate your electrons. If you move a single electron through $1V$ of potential the electron gains $1eV$ of energy where ...


1

My question is that if we repeat the above procedure in vacuum, would the sparks be produced? Not in perfect vacuum, because there is no gas to ionize. An understanding of this behavior can be attained by studying Paschen's Law (see http://en.wikipedia.org/wiki/Paschen%27s_law), which describes the breakdown voltage for given distance and pressure.


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


2

Take a capacitor and put it across a battery. There will be a transient current as the electrons go towards the anode . This happens very fast and the current is small. If you short the capacitor with a wire, the battery will empty all its charge on the short, which, depending on the battery can really be damaging. Your body accumulates some charge which ...


8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


2

The whole electrical power grid is connected to ground. I don't know the details of other regions, but if you are in North America, the two current carrying conductors in a residential electrical outlet are called "hot" and "neutral". The "neutral" conductor is connected to the Earth at many places. If your bare feet touch wet Earth, and your hand touches ...


1

If you are not in a complete electrical circuit, any electric shock caused by touching a charged object or wire is brief. These "static shocks" are slightly painful, but they are (rarely) dangerous or fatal. I'm sure you've experienced a minor static shock. By wearing insulating footwear, you break a complete circuit and forbid a flow of electricity from ...


0

Wouldn't the photons or waves of EMF just fly away into space and be lost (the energy would be lost, not stored) One must distinguish between electromagnetic waves and, e.g., static electric and magnetic fields. Essentially, ('real') photons are associated with electromagnetic radiation (radio, light, x-rays, gamma rays). Electromagnetic waves ...


2

Your argument that the energy should radiate away would be true if your inductor were a good antenna, in which case it would be a bad inductor! The problem is an impedance mismatch: The inductor produces a magnetic field (which stores the energy you inquire about), but little electric field. That is the wrong ratio, or impedance, to couple to the vacuum ...


0

A moved charge - in this case an electron - which is accelerated in a circle (the inductive coil) will induce a magnetic field. How does the electron induce the magnetic field? The electron has a magnetic moment and it spins. The movement of the electron in the coil align the magnetic moment and all moved electrons induce the common magnetic field of the ...


0

I think this is really about which way you count current and voltage to be positive. For every element in a network you can define a a current and a voltage. If voltage and and current point the same direction, it's called "receptor". If they are in opposite direction its' called "generator". The most common convention is to use "receptor" for resistors, ...


1

First let's establish the situation in which the result actually holds. Voltage itself is only well defined in electrostatics, and this result only holds in a steady state. In an ideal battery, there is no energy loss inside the battery during operation, and in the steady state just as much charge flows into the battery as flows out of the battery, and ...


1

The bleeder resistor is across the capacitor so they have identical voltage across.


3

Imagine a free-standing battery (not connected to any wires) and take a closed loop through the battery, out one terminal, and back in the other terminal. The total work done in moving a test charge around that loop must vanish. For this to happen, the change in electric potential outside of the battery must equal the negative of the EMF change within the ...


2

I recently answered a similar question here. The ideal capacitor equation $$i_C = C\frac{dv_C}{dt}$$ assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal. When you write $$iR = v_C$$ it is necessarily the case that $$i_C = - i$$ To see this, assume that both positive labelled ...



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