# Tag Info

## New answers tagged electric-circuits

0

Current is a measure of how many electrons go past a particular point in the circuit every second. So there are electrons rushing into one side of the light bulb, and rushing out of the other side. The number rushing IN each second is equal to the number rushing OUT each second. If that wasn't the case, then there'd be a build-up of electrons inside the ...

1

Under the scenario you presented, all you've done is made a bigger battery. Which is just sitting there, so there is no current flow. If you open up a standard 9 volt rectangular battery you will find six 1.5 volt cells connected in series, i.e. exactly as you described.

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Imagine a battery on a shelf, not connected to any circuit. That is for all practical purposes a circuit with infinity resistance. By the OP's logic it should self-drain instantaneously, and there would be no such thing as batteries.

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The only real answer is: in parallel, because if one bulb burns, the other one still burns. That's the answer chosen for a lot of daily-life items (your home, car etc) where, if one item fails (like the hairdryer), the others (like the light bulbs) should still burn. This is not really the answer yet to what you have asked for, because your question is like:...

1

Why does the current remain the same? Interesting question. As you could see from the examples with falling water or sliding blocks the gravitational potential is responsible for energy release in mechanical storage systems. For a battery this can't be the reason of energy storage. Kinetic energy inside the a battery can't be the reason too. At the end in ...

4

None of the above. Electrons are negatively charged, always. They do not become positively charged under any circumstances. In DC circuits they flow (or rather 'drift' at about 0.1 mm/s) only in one direction, from the -ve terminal to the +ve. In AC circuits they flow forwards and backwards in the wire, changing direction 50 times per second. They don't ...

8

I'm surprised that no one has yet mentioned the hydraulic analogy for electricity to help the OP understand better. A brief summary of this analogy is: Electricity is like water flowing through pipes. Current = amount of water flowing through pipe Voltage = pressure of water Power = water pressure x water flow (voltage x current) Resistors = constrictions ...

1

But if the amount of current flowing into the filament of the bulb = the amount of current flowing out of the filament and at the same time it is producing photons(light energy) [and some heat energy too] then aren't we creating energy ? Which is not possible. Current is the flow of charge over time , Q/t. It can be larger of smaller depending on the ...

1

If the amount of electrons remain the same, where does the energy come from? Batteries have energy stored inside. The energy inside a battery could accelerate electrons to a very high velocity. However, this doesn't happen in a circuit. Just as the battery speeds up an electron, the electron hits to another one and remains constant speed. So the battery ...

9

Imagine a block sliding down a slope and that there is an amount of friction between the slope and the block such that the block slides down the slope at constant speed. As the block slides down the slope it loses gravitational potential energy and an equal amount of heat is generated due to the friction between the slope and block. The block does not ...

5

If I have a circuit with resistance R and voltage V, I get a certain current - that's Ohm's law, $I = \frac{V}{R}$. Now imagine you have two such circuits - completely separate from each other. Each will have the same current. Let's say the voltage is 1 V, and the resistance is 1 A: Now if I connect the terminals of the two voltage sources together (...

0

Based on your drawing, with the batteries all in parallel, the resistor will have the same voltage across it whether you have one battery in the circuit or more than one battery in the circuit. Assuming that each battery produces the same emf, this will give you one value of current through the resistor, regardless of how many batteries are in parallel. ...

1

From the wording of the question I would assume the OP didn't want formulas or a very technical answer, so I'll attempt answering in layman's terms. What does resistance do? It resists the flowing of current. Given the same voltage, the bigger the resistance, the smaller amount of current can flow. Now, imagine that there is a resistor. You put another one ...

3

It seems like you get the rational answer but lack the ability to feel it: imagine you have 5 doors of different sizes and a thousand people to pass from a to b. If all the doors are in a row (so everyone has to pass every door), it will take a lot more time compared to the situation, where you place all the doors next to each other so that every person can ...

3

Let the potential difference $V$ is equal to sum of potential difference $V_1,V_2...$ In series connection $$V=V_1+V_2..$$ $$V=IR_1+IR_2...$$ $$IR=IR_1+IR_2...$$ Therefore the equivalence resistance, $$R_S=R_1+R_2...$$ ,which always greater than individual resistance While in parallel combination $$I=I_1+I_2+...$$ $$=>~~~~I=\frac {V}{R_1} +\frac {V}{R_2}... 11 Question: Which of two pipes of equal length offers less resistance to the flow of water, one of which has twice the cross sectional area of the other? Answer: The one of twice the cross sectional area. But the one of twice the cross sectional area can be thought of as two of the smaller cross sectional area pipes in parallel. This analogy gives an idea of ... 1 When you ask: So I was told in the physics class that the resistance in a parallel circuit is lesser than the resistance in a series circuit. This question only applies when two resistors are connected in series, versus the same two resistors connected in parallel instead. It's important to understand that apples-to-apples comparisons can only be made ... 8 Suppose you have a voltage V between two points A and B in a circuit. If initially you have a resistor of resistance R_1 between A and B, the current flowing through the resistor is I_1=V/R_1. Now if you connect another resistor R_2 in parallel to the resistor R_1, then the former will have the same voltage V across it (since it is connected to ... 2 The range of a given voltmeter can both be increased and decreased. We know that for converting a galvanometer of resistance G into a voltmeter of range V, a high resistance R given by$$R = \frac{V}{I_g}-G has to be connected in series to the coil of the galvanometer. In order to increase the range of the voltmeter, R has to be increased. It ...

0

Symmetry. At the left hand node the current can go through either the $2\Omega$ or the $4\Omega$ resistor and then the rest of the circuit, $\frac{2}{3}\Omega$, $2\Omega$ and $4\Omega$. At the right hand node the situation is exactly similar in that the current comes from an identically arranged $\frac{2}{3}\Omega$, $2\Omega$ and $4\Omega$, and then ...

1

If you're considering the following: then the resistance between the ends is $1 + \frac{1}{2} + \frac{1}{3} + \cdots$ which doesn't converge.

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it depends on how you connect the resistors if you arrange them in a grid, where they connect both going sideways along the stairs and up/down the stairs, then I think it would tend towards zero but if you have separate lines of resistors going up/down the stairs then it should be infinite?

0

I'm imagining this infinte staircase: +--+ --> | | +--+--+ | | | +--+--+--+ | | | | +--+--+--+--+ --> | | V V If the staircase has infinite underfill, by symmetry, the resistance of two points on the edge will be twice that between the same two points on the infinite ...

-2

Let me tell you some notable points these should help you solve such problems from the competitive exam point of view: =>A completely uncharged capacitor is like a wire, it offers 0 resistance =>A fully charged capacitor is infinite resistance and no current flows through it =>A battery that you have represented is an ideal battery and ideal batteries do not ...

-2

Your source voltage is a pressure of electrical charge, with a degree of resistance (obstruction) somewhere in-between zero and infinity, towards a lesser pressure of electrical charge. If the source pressure is not kept constant, by electrical input to the source, then this source pressure will diminish for as long as your current is kept constant. So any ...

3

While the paper linked to in sammy's answer does provide an equivalent circuit (page 5988), it seems of little practical use, since even in (unstable) equilibrium the current in L2 rises linearly with time. Within seconds (at most) it will reach a limit: saturation of the transformer core or maximum current (either I=U/R where R is the resistance of the ...

0

In a superconductor, the electric resistance is equal to zero. This is why an electric current can circulate forever in a superconducting ring even when the battery has been unplugged! This is how magnetic fields are created in MRIs. Source:- http://www.supraconductivite.fr/en/index.php?p=supra-resistance-supra

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I think the point made is that the kinetic energy of the electrons (or other charge carrier) will normally be far higher than $kT$. That's because although collisions with the lattice are frequent the electron loses very little energy with each collision. The point being made is that in a collision between a light object and a heavy one very little of the ...

2

The following paper is devoted to electrical analogues to mechanical oscillators. Fig. 6 illustrates a simple coupled LC circuit which imitates the inverted pendulum on a cart given in Fig. 5. It does not include a circuit to balance the 'pendulum' as in the video. https://core.ac.uk/download/files/508/12941830.pdf I also found the following which may be ...

-1

Potentials do not have retarded time effects. If electron is removed from the sun its effect on the potential on esrth is immediate . So the electronics will be likely tip top. The limit of speed nowadays in space is set by the rocket gas engines that they use. Basically they use the chemical bonds energy to accelerate and push gas outside.

1

My answer is r. 2 resistances on the vertical line are redundant as no current flows via that route due to the symmetry of the problem. Consider this circuit joined to the source with A joined to the positive terminal, then current will equally split along two possible routes from A. Due to symmetry of the problem, the electric potential energy above the ...

0

I asked my physics teacher, and the reason very little current (if any) is induced is that hysteresis in the core means it will quickly become saturated if the direction of magnetic flux does not change, and hence the secondary coil will not experience change in flux.

1

Resistors are in series if they are connected "end-to-end." In a series configuration, the current that flows through one resistor is the same as the current flowing through the other because that current has no where else to go. In this configuration, the voltage across the entire circuit is divided between the resistors. Resistors are in parallel if ...

3

Here is the question in context. I'm thinking it could be something to do with the internal resistances of the cells. Given the context, it seems reasonable to deduce that the person who wrote the question wanted you to consider the internal resistance. In real life, car batteries are made up of a series of cells. This fact makes this question ...

1

Edit: while my answer is indeed correct, RedGrittyBrick's answer properly addresses the context of the OP's question Nobody can explain why a battery made of a series of cells would not be suitable for a car battery, because such series of cells ARE INDEED suitable for car batteries. In fact, every car battery in the field today is a collection of 6 lead-...

3

If you remove all resistors the voltage drop will be across the wire. (Because the wire probably has a very small resistance the current through the wire will be very big and the wire will get very hot). if there are resistors in series connected by wires, the resistance of the wires is usually neglected. You can easily see that this is reasonable because ...

1

there is an electric field inside the wire, and there is a loss of potential energy, or voltage as they move but this drop in voltage is usually negligible (thought not in some applications) and we only consider that the drop in voltage comes only from the circuits elements o loads. This idealization often fails not with the wires, but within the battery ...

0

Let's first look at cases where this is not the case. When the lamps are connected to a constant current source, current is indeed "divided" over the lamps in parallel. Assuming equal resistance R, the bulbs will both see a current of I/2, and the dissipated power in each is I²R/4 or in total I²R/2. In the case of the bulbs in series, the current I will ...

0

Lets start with two light bulbs, 120v AC, 60W each. If you connect them in parallel to the mains, each receives 120v, 0.5A (= 60W). If you now connect them in series, each light bulb now receives only 60v (due to voltage division), and assuming the same current (0.5A), it only receives 30W! Therefore, if the light bulbs light up at all, they each only has ...

2

Your mistake in the above is in what you call the force required to move the charge from across the potential difference $V$. You identify the force as $\vec{F}=Q_{test}\vec{E}$, when it should be $\vec{F}=-Q_{test}\vec{E}$. This makes physical sense -- the reason it takes work to move the charge is because you need to oppose the electric field present in ...

1

When dealing with inductors Sears & Zemansky state that "we need to develop a general principle analogous to Kirchhoff's loop rule". With an inductor present in the circuit they state that there is a non-conservative electric field within the coils $\vec E_n$ as well a conservative electric field $\vec E_c$. Assuming that the inductor has negligible ...

-1

Since, the switch is connected to both the terminals of resistance by a wire and the wire must have negligible resistance so the both terminals of switch and resistance are at same potential or better to say that these are equipotential surfaces. . ------.1. .2.----- . | | . | | . 5------3///\4----6 . |. ...

0

The real part of the impedance $Z_{aa'}$ is resistive so all you need to decide as to what component, inductor or capacitor, you would assign a negative imaginary part to. Perhaps do it by a process of elimination or write down the general formulae for the reactances of capacitors and inductors?

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