# Tag Info

0

Since this is a homework question, I will point you in the right direction rather than do your work for you. Reading the question carefully, I believe you are trying to solve the following situation: There is nothing ambiguous here. You have two "red" charges (identical, 1.8 uC), and need to determine the value of the "green" charge such that the ...

1

One equation is for resistive circuit and the other is for capacitive circuit. Two can not be merged together.

0

I know that if all the resistors were the same resistance than the way to get the max current to go through them and hence the max voltage and power would be to place them all in series, No, you place them in parallel. For a resistance, the power is given by $$p_R = \frac{V^2_R}{R}$$ Since the maximum voltage you have is 30V, the maximum power is ...

6

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$\dot V=-\frac1\tau V,$$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...

1

Like any set of equations in physics, this is true when the conditions stated apply. Here, you are using two equations: $$V(t) = RI(t)$$ when the current passes through an element that is a perfect resistor, and $$C \frac{dV}{dt} = I(t)$$ when the current passes through an element that is a perfect capacitor. So, in a simple RC circuit consisting of a ...

5

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...

18

The identity $$V = K \frac{dV}{dt}$$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...

0

The key think you need to know is that if you apply a voltage $V$ and a current $I$ flows then the power dissipated is: $$P = VI \tag{1}$$ You know the power at 220V so you can calculate the current, and if we assume the kettle behaves as a simple resistor then it will obey Ohm's law so you can calculate the resistance. $$R = \frac{V}{I} \tag{2}$$ ...

4

I need to know why that is the answer. (Jump to the summary if you want to skip the details). As you may know, a transformer is essentially two inductors that share most if not all of the magnetic flux linkage. Recall the formula for the voltage across an isolated inductor $$v_L(t) = L \frac{di_L}{dt}$$ Now consider two inductors, $L_1$ and $L_2$ ...

0

I'm going to write a really simple answer... A short summary of what goes on in a step-up transformer: In an AC source, both the magnitude and direction of current changes As the primary windings of the transformer is connected to an AC source, there will be a change in magnetic flux (remember: the magnetic field strength around the coil is dependant on ...

1

The transformer operation is based on Faraday's Law $$\nabla\times\mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t}$$ This law relates the generated electric field (resulting in electromotive force in a circuit) with the variation of the magnetic flux density. Also indicates that a changing electric field generates a density also varying magnetic ...

0

Note that your circuit is equivalent to this:

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When I'm confused about voltages, I always remember how do you measure a voltage. You take two "needles", you stick them in any part of the circuit and you measure the potential difference. In this case $V_1$ gets measured by sticking the needles around $D_1$ and similarly with $V_2$ and $D_2$. $V$ is measured by sticking these needles in front of $D_1$ and ...

1

In your diagram, the devices appear to be connected in parallel and $V$ appears to be voltage across the left-most wire which must be zero. Since you've designated the devices with the letter $D$, I assume they are diodes but diodes aren't drawn as rectangles so I honestly can't write much more about this. As I'm typing this, I see a comment to the effect ...

1

Actually V=0! Notice that the left branch shorted the right one, because the points A & C are the same, thus Va = Vc.

0

Well, a net current results. I will write the general equation for the current, and determine all possible scenarios, keeping in mind that in writing this equation I assume the current is going clockwise (I assume the left voltage source is at a higher potential, and if this wasn't the case, my current will simply be negative and it will be flowing ...

1

A voltage or current given as a complex constant is a phasor. A voltage given as the complex constant $V_z$ represents the real voltage $$V(t) = \operatorname{Re} \left( V_z e^{i\omega t} \right)\ \ ,$$ where $\omega$ is the voltage's angular frequency and $t$ is time. Currents represented as phasors work the same way.

1

This is a parallel circuit, not a series one. So the currents not need be the same, but the potencial difference is the same instaed. In a series circuit the current is the same but the potencial is different in the many elements of the circuit. You can think about this like the current is a flow of water, since is parallel, the current (the flow) "divides" ...

1

The two resistors are in parallel. This means that at $A$ the current splits between them relating to their reistance. So the current throw the top resistor is $3\,A$ and throw the bottom resistor is $1 \,A$. If we use Kirchhoff's current law which states, that in any node (like $A$) the current flowing into the node is equal to the current flowing out of ...

3

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.

3

It is quite easy to use a step-up converter to generate almost any voltage. The question is "voltage at what current". The power a battery can deliver is finite (power = voltage times current), but you can convert voltages in many different ways. The most obvious is an oscillator (inverter) followed by a transformer and a rectifier, but there are more ...

1

I seriously doubt that the batteries were putting out 30 kV. You surely misread or misheard something. The chemistry of batteries is such that individual cells produce from a few 100 mV to a few volts. A 30 kV battery would require 1000s of cells, which would make no sense at all. In addition, 30 kV is much more difficult to handle and would be much less ...

1

Another useful analogy, apart from the gravity one described by David Z, is temperature. You can think temperature as your potential, and the heat flow as your current. Two points of space may be at different temperature, but if they are correctly insulated, they won't exchange heat. The heat will flow only if they are connected somehow. For the current is ...

3

The analogy of electricity to flowing water may come in handy here. In this analogy, a potential difference is like a difference in height. One lake on top of a mountain and another in a valley, for example, might represent the two terminals of the battery, which are at different potentials. If you think about that situation, it's clear that no water flows ...

0

If you insert a dielectric in a circuit, you will not see any current but obviously there is a potential difference across the dielectric. To have a potential difference, you just need an electric field inside the material. This electric field might drive a current if the charges are mobile.

1

You are increasing the number of electrons in the wire, but only by a very small amount. There's a somewhat clichéd but still excellent analogy for electrical circuits called the hydraulic analogy. In the hydraulic analogy the power supply is a pump, and the pressure is the voltage. The water represents the electrons, so the pressure generated by the pump ...

0

No. The voltage is the chemical potential of photons instead to relating to electrons, in other words, an electromagnetic field is established across the wire which moves the electrons. Back to your question, the number of electrons does not change in the wire at the open-circuit scenario, aka the electrons in the wire can't flow into or from anywhere.

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Voltage is potential difference. A higher voltage means that there is more energy that can be used from the same amount of current. In effect, increasing the voltage is roughly analogous to adding more potential energy per electron, as opposed to current, which is moving more electrons through the wire.

4

You make a good point. In a battery a chemical reaction (a redox reaction) creates the potential difference, and the potential difference is calculated assuming the electrons start and finish in well defined energy states. If electrons returning to the cathode have some residual kinetic energy then the could affect the reaction and change the EMF. However we ...

2

You're quite correct that there will be some current flowing, so there must be a voltage drop due to the internal resistance of the battery. The EMF measured by any voltmeter will always be less than the true EMF. If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be:  V = ...

1

Of course lightning "has a frequency component". If it didn't have non-zero frequencies, it could never start or would last indefinitely. Lightning is a huge current pulse over a short time, usually several pulses over a few milliseconds. But more importantly, the current is started and stopped very abruptly, which by necessity means it has a broad ...

1

According to wikipedia, average duration of a lightning is $30\,\mu s$. If we take a gaussian current splash with $\sigma=30\,\mu s$, its spectrum will be a gaussian with $\sigma_k\approx33\,\text{kHz}$. This doesn't look like DC.

1

It is very hard to use a voltage source to induce charge on an insulator. The reason is that by definition, an insulator does not conduct electricity - so if you apply an electrode at one place, you will not move electrons elsewhere, and so you cannot induce a net charge (the best you can hope for is to create polarization, and maybe pull off a handful of ...

1

Voltage has absolutely nothing to do with charge. I can "move" an infinite amount of charge trough a superconductor with zero voltage. Are you asking about the relationship of charge to voltage on a capacitor? That's a linear relationship: Q=C*U. The charges, in that case, are not "created" but merely separated. If you want more charge for the same amount of ...

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