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Actually the shunt resistor has low resistance than the ammeter resistance. so most of the current will go through the shunt resistor(because of its low resistance) and only a small amount of current will flow through ammeter. then how the ammeter will damage?. According to me, the shunt resistor has low resistance so most of the current will flow through ...


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Because an ammeter has low 'resistance/impedance'. It is designed to measure the current. A low resistance increase the current passing through that wire. When an ammeter is placed in parallel almost all current flows through ammeter and not through the wire and the resulting current will 'blow a fuse'. Conversely, a voltmeter has a high impedance and is ...


1

As the value of $b$ increases the resistance between the outer and the inner shells will converge to $1/4 \pi \sigma a$. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be $1/4 \pi\sigma a$. We can think of the situation in which there are two shells in the infinite sea of poorly ...


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In series the current has to pass through the shunt resistor as well as the ammeter. If the current rises then the voltage to drive the current through the shunt resistor increases ($V=IR$) - this will reduce the voltage available to the rest of the circuit and the current will drop... In parallel the ammeter has a lower resistance than the shunt resistor ...


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But isn't $I^2R$ equal to $V^2/R$, therefore if $R$ is constant doesn't the power depend on the square of the voltage so surely it doesn't matter whether it is high voltage or high current. Consider the wires connecting the power plant to the appliance; let the effective amplitude of oscillating current flowing through all wires be $I$ and let $V$ be ...


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If you keep the resistance constant, then $V=IR$ means that voltage is directly proportional to current. If you keep the power constant, then $V=\frac{P}{I}$ means that voltage is inversely proportional to current. However, because $V=IR$, we can write that $P=I^2R$. Therefore, if we say resistance is constant, then power must change with current, which ...


1

for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


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If you are considering a fixed charge on the capacitor, it won't matter at all. The charges have plenty of time to distribute themselves evenly over the plates, so there is no current anywhere, so no voltage drop. If you are considering conditions during the charge cycle, the charges in the low resistance plate will spread out more quickly. If the leads ...


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The thrust a rocket can generate is proportional to the mass flow (thus $\frac{dm}{dt}$) and the velocity at which this mass leaves the rocket, often called the effective exhaust velocity, $v_e$. So the sum of all forces acting on the rocket, when also neglecting the fictitious Coriolis and centrifugal forces and assuming a pure vertical ascent, will be ...


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Technically, you can maintain a constant (or near-constant) temperature of the conductor in spite of Joule heating by cooling it with temperature regulation. Practically, resistance typically does not change dramatically under moderate Joule heating.


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Without going too deep pick for example the first circuit (from left to right) and call the first resistor $R_1$, the other $R_2$, same with the capacitors. Remember the definition of capacitance: $ C = Q / V $, and how its charge is related to the current through it: $ i = dQ / dt $. Then label the currents, for example call the input current $i$, then by ...


3

The traditional least-squares fitting or chi-squared minimisation route of fitting a straight line makes the implicit assumption that the errors on the x-axis quantity are negligible. If that is so, then there is no reason why you can't use the uncertainty in the gradient as the uncertainty in $R$. I guess from your question though, that this is not the ...


1

NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.


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No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.


1

Well, yes you can, but it is usually very hard. Here are the steps: Solve the Laplace equation: $$ \nabla^2V = 0 \, .$$ In your case, find the general solution in spherical coordinates. Try to use every simplification you can. You might wonder why you don't solve Poisson's equation: $$ \epsilon\nabla^2V = \rho \, .$$ That's because a conductor is an equal ...


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The general derivation of Pouillet's Law is given in https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance


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The electrical resistance of a sphere is calculated in https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance


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As pointed out in some comment, electrons are being accelerated in the process of charge. This generates electromagnetic radiation. Try doing the calculation using Poynting vector, the way Maxwell defined electromagnetic energy.


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Pouillet's Law is valid in general, but the specific values of l and A depend in the shape and environment of the resistor. See https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance


2

The voltage V that you're computing would be induced between the upper and lower sides of your orange block, and $L$ in your computation is the distance between these two sides. This voltage is induced between any two opposite points on the upper and lower sides provided they are in the area that is covered by the magnetic field. If you connect a wire with ...


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In regard to the first, we have a 5v supply, which from my understanding means that if you were able to enclose a coulomb of charge eminating from the negative terminal, you'd find that it has 5 joules of energy. This isn't the typical understanding. A 5V (ideal) supply (source) maintains a 5V potential difference across the terminals independent ...


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First, a couple of asides: A. When circuits are drawn, the convention is that straight lines represent perfect conductors. Any real resistance is shown as a resistor, and hence, gets included in any IR drops. You don't go and say "well, there are some other resistances." If they are important, they are shown. Circuit analysis of diagrams is done with this ...


0

I'd like to start with a slightly pedantic aside: I think you should try to avoid the phrasing of electrons "having energy"; what they have is potential energy, and potential energy is relative. This is important to remember. Question 2 is trivial, so I'll answer that first: there is no such thing as zero resistance, so the wire from R3 to the batter's ...


0

Power in a circuit is defined by P= I x E P is power in watts or joules per second I is the flow of electrons in coulombs per second or amps E is the electormotive force or voltage the energy of the electrons is lost to heat your first question the power supply see only the circuit as a whole it dose not see the individual resistors for R1 the total ...


1

You have to think more carefully about what exactly $Q$,$I$ and $t$ signify: In $$ Q = I t$$ $Q$ is the charge that is transported by the current $I$ during the time $t$. If you now write $$ \frac{I}{Q} = \frac{1}{t}$$ then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...


1

Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ...


0

This is how ideal and impossible circuits elements behave, but it's a starting point for a simple analysis: At discontinuous changes in circuits, 1) inductors have the same current immediately before and after the discontinuity, but can have discontinuous voltage changes. The current will then change exponentially/sinusoidally/both. 2) capacitors have the ...


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if we consider a current of 6A flowing through circuit we will then get a equation E=6I*R 6IR = 5Ir = 5/6 ohm total current is 10/(5/6)= 12A


1

A metal conducts quite well because the there is an electron band that crosses the Fermi level. So, electrons can easily be excited to increase their momentum a bit and consequently move in one direction. Now if you add one electron to the wire, the Fermi level rises. However, you would not be able to see the increase caused by one single electron (or a ...


0

Adding extra electrons isn't going to make any difference, at least as a first approximation. Suppose we have a wire with some applied voltage $V$ and some current flowing $I$. What this means is that the power supply is injecting $I/e$ electrons per second into one end of the wire and extracting $I/e$ electrons from the other end of the wire. The ...



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