New answers tagged electric-circuits
1
Hey you're getting wrong, the equivalent emf(voltage) of system will be 16V-8V because both have opposite poles facing each other, so their will be net flor of current (-ve to +ve) according to the cell of greater emf(16V cell).Then your total resistance is $$5+1.6+1.4 = 8 \Omega$$(all are in series) . $$I(Current) = E(e.m.f or Voltage)/R(Resistance) = 8/8 = ...
1
The voltmeter does indeed draw current, but very small. Its internal resistance should be high, unless it's a very cheap voltmeter.
Putting the voltmeter's internal resistance, let us say 1MOhm just to make up a value, in series with the battery's internal resistance of a fraction of one Ohm, makes a voltage divider providing practically the entire ...
2
why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance
It does, but the voltmeter probably has an impedance of 10,000,000 ohms - as most multimeters do - so the current flowing is negligibly small (0.6 microAmps) and therefore the voltage drop due to battery internal resistance is ...
1
You should not connect different batteries in parallel.
If you do, the battery with the highest voltage will discharge into the other one, until they end up with equal voltages. If the second battery (the lower voltage one) is a rechargeable, then it will be charged by the first one, again until the two have the same voltage. In this case the end voltage ...
2
In ideal circuit theory, the parallel connection of two voltage sources results in an inconsistent equation, e.g., a 3V and 2V source connected in parallel, by KVL, gives the equation: 3 = 2.
In the real world, batteries are not ideal voltage sources; batteries can supply a limited current and the voltage across the battery does, in fact, depend on the ...
0
The answer is YES, it is possible for current to flow in an open circuit. The only requirement is that the current be "alternating" current. A capacitor is essentially an open circuit, and alternating current will "flow" trough it.
0
This isn't what you're asking, but it is possible for currents to flow in an open circuit. Even a plain old block of copper is an "open circuit", for instance, but you can induce eddy currents to flow in it by applying a changing magnetic field. In a regular circuit this effect would be very small, though, as the copper wires have small width.
1
Let the resistance of the original wire be R.
R = ρ (L/A)
Now l = L/5
R’ = ρ (L/5A)
Or R’ = R/5
Now 5 resistors of R’ are connected in parallel
1/R(net) = 5/R + 5/R + 5/R + 5/R + 5/R
or
1/2 = 25/R
or
R = 50 Ω.
0
You're confusing charge with voltage/potential. Objects have charge, voltage is measured between objects. If two objects have no charge, the voltage between them is 0. If two objects have the same charge, the voltage between them is still 0.
Voltage is always a relative difference between two points. When we say "Terminal A is at 5 volts", what we ...
4
A human body may reflect and absorb radio frequencies, though not very efficiently. It may as well act as a resonance chamber for certain frequencies. For a signal of 100 MHz, the involved wavelength is 3 m, and so it is possible that parts of your body are acting slightly as a resonant chamber. (for an optimal resonance, you should have 1.5 m diameter, too ...
1
The net charge of any of those internally connected pairs of plates is always zero. That is, when you charge the capacitors, charge doesn't leave the wire between C and D, it only moves along it, and is held in place by the electric field of the adjacent plates. If a circuit is completed that allows charge to flow from D's negative plate to A's positive ...
2
When you touch a hydro wire, and ONLY a hydro wire -- nothing else -- there's no potential difference across your body. Your entire body is at the same potential, which is that of the hydro line (often thousands of volts). This is why birds don't get electrocuted if their feet touch only one wire. Plenty of birds, particularly eagles, have been fried because ...
10
AC or DC, you only get electrocuted if current passes through your body. (Current passing through any part of your body can be dangerous, and possibly cause an electrical burn, but current passing across your heart is the one that's really dangerous.) Touching just one wire at a time gives the current nowhere much to go. You are right to think that some ...
-3
electric current always takes the way which has the lowest resistance.
When the birds aren't grounded the electric current won't take the way through the birds.
2
Your reasoning is correct. And as a bonus: to measure the resistance of the voltmeter, you put it in series with the ammeter.
-2
The wires have resistance. You had 5 resistors in series. If connected in parallel, they give a known parallel resistance.
1
Resistor is anything that pose resistance to flow of charges in a circuit.
Here is how it looks like:
$$R=\rho \dfrac lA$$
Where $\rho$ is material property, $l$ is length and $A$ is cross sectional area.
Here is how 5 resistors in parallel look like:
and the circuit diagram showing the same:
In parallel configuration voltage drop across each ...
2
Seems like you're not making the connection between the actual physical setup and the equations. So, here's a translation:
1) A length of wire is a resistor. By resistor what we mean is that when we apply a potential difference $V$ between the two ends (like from a battery) the resulting current $I$ is given by $V = IR$ where $R$ is the resistance.
2) The ...
0
The formula $V=\frac{Q}{C}$ gives the amount of charge that is there on one of the plates. The total amount of charge on both the plates taken together is zero! Both of them are oppositely charged.
1
For flow of charge, the circuit should be closed. In open circuit, no charge flows. If we connect both the capacitor plates it makes closed circuit, charge flows in the circuit, as a result charges on the plates neutralizes to zero.
If only +ve plate of the capacitor is only connected to ground there is no closed circuit. no charges flows from the ground.
...
0
Net charge on capacitor is always zero because there is equal and unlike charges on plates.
Hence capacitor is not charge storing device. It is electrical energy storing device.
In any form of capacitor, stored charge when charged by voltage V is q=cv
where +cv is stored in one plate and -cv is stored in another plate.
In this question charge stored should ...
0
Your question is the result of a poor wording choice by the Wikipedia author. While it may be possible to create a digital computer that makes use of AC power, none do. The power supply in modern desktop computers use a transformer and rectifier (and some capacitors for smoothing) to provide a constant, smooth $12\, \mathrm{V}$ DC current for the computer ...
1
They are different because in a transmission line we have distributed resistance, capacitance, conductance and inductance (meaning that each tiny segment of transmission line has its own tiny resistance, capacitance, conductance and inductance) while in RLC circuits we have lumped resistance, inductance and capacitance. Also RLGC doesn't model a transmission ...
1
Your argument is correct that there if there is no current in the resistor which is in parallel with the $2C$ condenser, then the charge on the $2C$ condenser must be $0$. This, as you probably already know, is because the two elements are in parallel and so they must have the same potential across them. However, you should give a clear argument as to why ...
0
Attention that
1.X is the impedance of the inductor
where X is the impedance of the inductor
2.The source is an AC, so you have to calculate the effective voltage it provides, which is to
divide square root of two from the MAXIMUM voltage amplifier. In your formula V=IZ, V should be calculated in this way.
1
Power formulas for DC circuits is not correct in AC circuits unless you use root mean squared voltages and/or currents. So $I_{RMS} = 0.87A$ and $V_{RMS} = V_{MAX}/\sqrt{2} = 389V$ and the impedance is $Z = {V_{RMS} \over I_{RMS}} = 447 \Omega $
Alternatively you could have computed maximum current from RMS current and find the impedance with maximum ...
0
But Va−Vb=−2ε ... how do I make sense of this?
It's incorrect to write $V_a = \varepsilon$. The voltage $\varepsilon$ is across the battery.
Try this: place a ground symbol on the wire between the battery and the $a$ terminal; this is your zero node or the place you put the black lead of your voltmeter.
Now, if you place the red lead on the terminal ...
0
Think of the open as infinite resistance. Your current will be $I = E/(R+\infty) = 0$. Now, voltage dropping across every resistor is proportional to the current and resistance: $V(r) = I * r$. Since infinite resistance is infinitely larger than R, all $E$ will drop across $\infty$ and 0v is left to R. Particluarly, $V_R = 0\mathrm{v} * R = 0\mathrm{v}$. ...
0
Ok , Potential is work done by electrostatic force as you move from A to B per Couloumb. So transfer 1 coulomb from A to B and as you move through the battery, E is directed from +ve plate to -ve plate . So work done by E is ($V_- - V_+$) , which means $-EMF$.
And yes that is an assumption that charge must reach the positive plate with the same velocity ...
3
I'm not sure I understood all your points. I suggest you to read this beautifull paper
Romer, R. H. (1982). What do “voltmeters” measure?: Faraday’s law in a multiply connected region, American Journal of Physics, 50(12), 1089. http://dx.doi.org/10.1119/1.12923
if you can find it. It's seems it's exactly what Prof. Levin is doing in his lecture. The ...
0
There are actually two slightly different versions of Thevenin's theorem. I think what you are describing is the weaker of the two: you can replace any circuit with a single voltage/current source and a single resistor. That version holds for any two-terminal network made up only of voltage/current sources and ohmic resistors. It fails as soon as you add ...
1
$\Gamma_{ii}$ is the $i$-th entry on the diagonal of $\Gamma = L^+ = (D-A)^+$, $\Gamma_{jj}$ is the $j$-th entry, and $\Gamma_{ij}$ is the entry located at row $i$, column $j$. Thus $\Omega_{ij}$ is a scalar, but you could assemble all such values into a matrix $\Omega$ that gives the resistances between all pairs of vertices.
1
Outside a current carrying conductor, there is, in fact, an electric field. This is discussed for example, in "Surface charges on circuit wires and resistors play three roles" by J. D. Jackson, in American Journal of Physics -- July 1996 -- Volume 64, Issue 7, pp. 855 . To quote Norris W. Preyer quoting Jackson, "Jackson describes the three roles of surface ...
1
(1) $v(t) = V_m\cos(\omega t + \phi) = Re(V_me^{j\omega t}e^{j\phi})$
(2) $\dfrac{dv}{dt}= \omega V_m\cos(\omega t + \phi + \frac{\pi}{2})=Re(\omega V_me^{j\omega t}e^{j\phi}e^{j\frac{\pi}{2}})= Re(j\omega V_me^{j\omega t}e^{j\phi})$
By inspection, comparing the last term in (1) and (2), the phasor corresponding to $v(t)$ is $V_me^{j\phi} = \vec V$ while ...
1
To put Zetta's answer in more visual terms: A battery "pushes out" electrons on the negative pole while "sucking in" electrons on the positive side. In the diagram you draw the 9V battery pushes electrons way more strongly than the 3V battery, so they move from the 9V to the 3V.
An analogy with tilted boards may also help: Suppose you tilt a board to the ...
1
As Brandon said above, the reason is that since one battery is of a greater voltage than the other, they do not entirely neutralize one another to the point where there is no current. To demonstrate this a bit more rigorously, consider the current produced by each battery. This analysis is possible through the principle of superposition:
The current ...
0
For a given circuit in a given technology, power increases at a rate proportional to $f^3$ or worse. You can see by looking at the graph in @Martin Thompson's answer that power is superlinear in frequency.
$P=c V^2 f + P_S$ is correct, but only superficially so because $f$ and $P_S$ are functions of $V$ and $V_{th}$ (the threshold voltage.) In practice ...
1
That equation is true for electrostatics. Inside an electrically neutral current-carrying wire, the electric parallel to the wire is zero. So outside the wire it's also zero.
More importantly, Gauss's law will tell you that the components perpendicular to the wire must also be zero.
So the electric field is zero everywhere for an electrically neutral ...
0
EMF is totally associated with a battery. In other words, it's the voltage produced by a battery. So, yes. It's the amount of work done in moving an unit charge. But, from the positive terminal to the negative terminal of a battery unlike potential difference.
Induced emf is not totally different. But, it's somewhat different because it's very unlikely to ...
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