# Tag Info

0

Second question answered first: in context of physics or electrical engineering, a battery is modeled as something that provides a voltage difference, a difference in electrical potential, that can keep that voltage (nearly) no matter what you do to it. So, some charge traveling (== some current flowing) will have no effect. Real live batteries are ...

0

About the first question: The potential is created by the charges themselves and is found from their spatial distribution. It does not mean anything to say charges loose potential even though their energy can be deduced from the it. Moreover, The potential can be found by solving Poisson equation as illustrated in the following example: Suppose 1000 ...

1

Yes, there is always an electric field around a battery, even when nothing is connected to the battery. However, if there isn't a current flowing, the electric field doesn't have much of an effect, because no energy is being expended doing anything interesting according to Watt's law, $P=VI$. The field does exert a Lorentz force $F=qE$ on a charged object ...

0

A battery is basically an electron pump. Inside the battery a chemical reaction pumps electrons from the positive end to the negative end. As the electrons move they create a charge separation that opposes the pumping action, so if the battery isn't connected to anything only a limited negative charge can be built up on the cathode and a positive charge on ...

1

The fundamental problem is that the Ohm's law is a simple mathematical approximation, neglecting the V-A characteristic of a real source, resistance of wires, or even nonlinear physical response of materials under extreme conditions. Using this approximation outside of its scope leads to a "0/0" problem, which is obviously wrong. Some other physical ...

2

Even if you put a superconductor across the terminals of a voltage source the current would be finite as all real voltage sources have a resistance. A circuit with a voltage source with no resistance does not exist.

1

The reason to use the alloy is because it has a much higher resistivity than copper and so the alloy wire will have a higher resistance which with standard laboratory apparatus can be measured more accurately. I also seem to remember that the temperature of coefficient is lower for some of these alloys ie for a given increase in temperature the resistance ...

0

The choice of a non-metallic sample is to get the resistance high enough to measure. The resistance of the rheostat, the voltage supply, and the hookup wires can make the measurement inaccurate if you do not account for them. If they are much smaller than the resistance of the sample they will not matter much. You also will not try to draw too much current ...

1

In such circuit see that all resistance at rim have same potential, since no resistance reduced it. At rim all have same potential as that at $a$. At center all have same potential as that at $b$ So, all have same potential difference $V_{a} - V_{b}$ Same potential difference means that they are connected in parallel.

1

If you did not have resistor R in the circuit the voltmeter would always give the same reading - the voltage of the battery. A thermistor changes its resistance when a temperature changes. You are not given a resistance meter. All you have is a battery, a resistor R and a voltmeter. The circuit as set up is called a potential divider which means that the ...

2

The helicopter and the power lines are at different potentials, the difference being so great as to cause the air in between to become a conductor. If you applied such a potential difference across a line worker it would probably result in death. You will note that the line worker is holding a metal stake which has a "pointed" end. This increases the ...

0

I think that you have faced a conceptual problem because you have used the relationship $V=RI$, rather you should have been using the definition of resistance $R = \frac V I$. Then as $V$ goes up and $I$ goes down the value of $R$ increases. Also you must be careful about $I=0$ (current equals exactly zero) and $I\rightarrow0$ (current gets closer and closer ...

0

Bear in mind that infinite resistance is equal to an open circuit. In real life if you have a battery not connected to anything, there will always be a voltage across the terminals. Even a lead-acid car battery will produce sparks if you short the terminals!

-1

At R→∞ voltage won't be 0 but little less than EMF off cell, since a series circuit with voltmeter and battery is complete. Yes, voltage across resistor reaches 0.Voltmeter reading is not EMF because it draw some current which pass through internal r and gets some potential across it. An Ideal voltmeter cannot tell you voltage across resistance in ...

0

-First, I'm trying to grasp the concept of why and how the voltage drop at the terminals of a battery depends on the resistance of the circuit. The EMF, $\mathcal{E}$, exists chemically and is always present. If there is a path for current to flow, then there will be a voltage drop across the internal resistance, $r$, and the external resistance, $R$. ...

0

Assuming steady state means that you have no current flowing in the branch with capacitors in it. So think of the circuit first as a potential divider made up of the battery and the two resistors. From this you can find the potential difference across the two capacitors.

2

You have two basic options: Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or Use Kirchoff's laws to derive the equivalent resistance. If you choose option 1, I'll help you out by revealing the parallel resistors in this ...

0

In a perfect transformer, if we run a current through either the primary or secondary coils, we are guaranteed in the quasistatic case that $\Phi_1=\Phi_2$ for each individual turn of the coils. Now suppose we ran a current $I_1$ through the primary coil, and waited long enough so that $I_1$ was steady and $I_2$ (current induced on second coil) was zero. we ...

0

In practical terms, resistors in electrical circuits can be used for a wide variety of applications. If we put a small resistance in series with a conductor in a circuit, and know the value of the resistance accurately then we can measure the current through that conductor by reading the voltage drop across the resistor and using Ohms law to calculate the ...

1

Short answer: "neither". A resistance, in a circuit, causes a voltage drop for a given current. Alternatively, if the voltage is given, it causes a particular current to flow. As soon as a circuit gets a little bit more complex (more than one battery-resistor-light bulb-switch), it becomes completely impractical to use lots of batteries to control ...

2

An electromagnet has a lot of turns of wire, which gives it a property called inductance : the property of resisting a change in the current. You may be familiar with Lenz's Law $$V=-L\frac{d\Phi}{dt}$$ which says that an inductor with inductance $L$ will generate a back e.m.f. $V$ when the flux $\Phi$ through it changes. Now if you drive a current through ...

2

I think that this is a very good question because it makes one think beyond a "standard" explanation. When you study electromagnetic induction you learn about the magnetic flux change through a close loop which produces an induced emf. However the loop does not have to be a conducting loop. If it was an ideal dynamo with no resistance, friction etc no ...

0

Power factor means ratio of power consumed by LCR circuit to power consumed by resistive circuit. If Power factor is 0, it means No power is consumed by LCR circuit, although power supply is given.It acts like open circuit. When power factor is 1, it means all power supplied by source is being consumed. This is called resonance condition. Power factor ...

-1

Google translate is a wonderful thing: Cut and paste the solution (in parts) into Google translate. It will not be a perfect translation but it should be good enough to provide an explanation as to how to solve the puzzle. For example here is the first paragraph: 20th edition , the role of VI . 3 ... čtverákčtverec ( 4 points , average 2.73 ; dealt ...

3

When you have a capacitor, current flows even though the "circuit" is not complete. This is because it's possible for electrons to bunch up - temporarily - in a conductor and generate a corresponding electric field. That is what happens in an antenna. An antenna is really a combination of an inductor (a straight wire) and a capacitor (when you put a net ...

0

I would focus on the current in the middle ($I_S$). When the circuit is open $I_S=0$ $$I_A=I_B=\frac{V_A}{R_A+R_B}+\frac{V_B}{R_A+R_B}=\frac{V_A+V_B}{R_A+R_B}$$ When it is closed you have $I=(V/R)_A-(V/R)_B$ and this current either adds or subtracts from the current across the bulbs. If $(V/R)_A$ is greater than $(V/R)_B$ then you will get $$... 0 Yes you are right. We have made an assumption that (charge flown in) = (charge flown out). There will surely be an accumulation of charges at least in some part of elements.In the case of accumulation of charges we can find the other parameters like Voltage drop etc., using Maxwell's equations. But by making this small assumption we have greatly reduced all ... 0 This is a very interesting question which has at least two answers. The first answer is for an electrical/electronic engineer who wants to get a correct answer by using any systematic and coherent method. The Associated Variables Convention is used. You assign a current direction to each circuit element and then assign a plus sign at the point at which ... 0 It is possible for a capacitor to have a net charge. However, most components have a small capacitance with respect to infinity. Unless you're dealing with a huge metal ball, it will take a large voltage in order to get any appreciable charge there. 0 emf induced by inductor is negative but emf applied by battery on inductor is positive. So, that inductor tries net voltage across it is 0. Emf applied by battery is E°sinwt of which LdI/dt is applied across inductor, RI across resistance and Q/C across capacitor. -LdI/dt is induced across inductor and not applied. Now, consider direction of emf is ... 0 1.) When you first close a circuit, there is a very brief period of time in which the electrons push each other forward "one by one", so to speak. This very brief period of time is probably on the order of nanoseconds, so we usually ignore it. After that, a steady state condition is established, and the electrons move all at once ... more or less. Don't ... 2 You've asked some really good questions here. Before starting, I want to first mention that the traditional picture of particles moving through a wire in electostatics is missing some physics; for instance, it ignores the quantum mechanical nature of electrons. The reason we still teach this model is because it captures the main effects (the phenomenon of ... 0 When you find the response of a linear circuit for a frequency \omega you are taking the differential equations that describe it and looking for the homogeneous solution. The true response of the circuit is this homogeneous solution plus a particular solution which depends on the initial conditions. That is the missing term you found. The particular ... 0 If you ask this question in full generality -- i.e., is it possible for any value of R1 to R9 to infer their value from the resistances between A..F -- then answer is no. Just imagine R_1=R_2=R_4=0. You will never be able to infer R_5, as there is a short circuit. Regardless of that, I doubt that there is generically a unique solution, since the ... 0 By connecting A with B and E with F you can reduce the problem to a 2x2 matrix. Same for all other combinations of two (or n-1 lines in general). I would then call the resistors R_{11} to R_{22}. If you short A with B and E with F, you combine resistors in such a way that R_{22}=R_9. Shorting out a horizontal line with a vertical line in the simplified ... 0 Input resistance and output resistance are parameters to help you use a particular circuit. In your example above the attenuator starts at Port 1 and finishes at Port 2. Removing the load resistor R_L for the moment suppose you were asked what the input voltage to the attenuator V_i was with the given supply. To answer such question you could reduce ... 0 I know that input resistance is the resistance as seen by the input terminals As written, this isn't correct. The input resistance is the resistance looking into the input terminals. Conceptually, this means that if one changes the voltage across the input terminals (only), the input current changes by$$\Delta i_{i} = \frac{\Delta v_{i}}{R_{i}}$$... -1 I don't really remember the voltage laws, but I can tell you with only one source positive current always flows from plus to minus on a voltage source so Ia and Ib must be positive as youve drawn the arrows, and Ia is correctly negative relative to the "+" and "-" you've drawn on the 21kOhm resistor. I recommend not getting stuck up in the laws and ... 0 Kirchoff's laws say that the voltage on the bottom wire is some constant V_0, the voltage on the top-left wire is V_0 + \mathsf{24~V}, and Ohm's law gives us that the voltage on the top-right wire is$$V_0 + \mathsf{24~V} - 12 ~\mathsf{k\Omega}\cdot(I_a + I_b).$$We can arbitrarily choose V_0 to be \mathsf{0~V} or \mathsf{-24~V} as you wish, but we ... 0 first of all when electrons are moving they are really similar to Newton's Cradle.the back electron pushes the front one. different in electric potential of terminals do work on electrons and causes circuit.now moving electrons as you said collide atoms which increases temperature. in fact potential electric energy turns to kinetic energy and kinetic energy ... 0 You don't need to supply any extra energy to return electron to + ve plate . Electron already have enough KE even after passing all the acquired energy by battery to resistor. Never forget that actual speed of elections is very huge. When these electrons were in battery they already had enough KE. When you applied resistor it just increased its energy. ... 1 To understand this better take the fluid mechanics analogy.Consider a pipe to be a conductor and the pressure across it to be the voltage difference and water flow to be current.Now consider a situation where there is water flowing from through the pipe,it is a matter of fact that if water is flowing there has to be some pressure difference between the ends ... 1 The currents will be the same. Using Kirchoff's current law, the total current going out of any node or any closed contour is 0. Suppose you have two resistances connected in parallel between nodes A and B. Draw a circle around the parallel resistances. The total current going out of the circle is 0. So outgoing current at A + outgoing current at B = 0. In ... -1 Looking for something else accidentally found your conversation Probably is useless after 3 years but let me make a comment Nothing new, but putting it in on other way Keeping the current fix in the formula Q=I2Rt, shows that the energy amount converted to heat in a resistor is proportional to the resistance. This is because in order to keep the current ... 1 When you analyze a circuit, you have to pick a convention for your potential and current flow. It doesn't matter whether you pick left-right or right-left, up-down or down-up, you just have to pick. After that, you do your calculations. If you drew the current "to the right" and you end up with a negative number, you know the (positive) current was really ... 2 The Key to your question is the distinguation between ideal and real sources of voltage and current. If we stay at the batteries: At the moment you attach batteries paralell to your body, what happens is that one part of the body has another electric potential than the other one, thus (Ohms law) a current will flow in your body. At this point the current ... 2 The powerline can supply large amount of current, whereas the doorknob cannot. Each power supply comes with voltage and max rated current. Let us say body resistance is 5M ohm. Then current flowing through the body when you touch powerline is 20k/5M = 4mA. Now powerline can easily supply 4mA of current. But the doorknob cannot supply 4mA. So the actual ... 2 A capacitor can contain a certain amount of charge for a given voltage:$$Q = CV When you have more than one capacitor in parallel, they have the same voltage (because they are in parallel), and each stores a certain charge. The total charge (at a given voltage) will be the sum of the charges on all the capacitors. Now if you have a certain load (for ...

0

The live wire oscillates and either drains or supplies electricity to the neutral wire. The live wire oscillates between 230V to -230V and the neutral wire always stays 0V.

1

You can confine an electric field in a Faraday cage, if you want to, then the volume is finite. – CuriousOne

Top 50 recent answers are included