# Tag Info

## New answers tagged current

3

\newcommand{\p}{\partial} \begin{aligned} \p_{\mu}J^{\mu} &= \gamma_{\nu}(\p_{\mu}\p^{\nu}\phi)\gamma^{\mu}\psi+\gamma_{\nu}\p^{\nu}\phi \gamma^{\mu}\p_{\mu}\psi-m\p_{\mu}\phi\gamma^{\mu}\psi - m\phi\gamma^{\mu}\p_{\mu}\psi\\ \end{aligned} Now the first piece can be written as \gamma_{\nu}(\p_{\mu}\p^{\nu}\phi)\gamma^{\mu} \psi= ...

0

Take a battery of 4,5 V. Take one light bulb and some wires and arrange a simple circuit, measure the current, and lets say it is 2 A. Then, take another light bulb and connect it in parallel to the first one. Overall current will now be 4 A, it will increase. The rule is intact, as you can see, because in each branch now you have 2A which ads up to total of ...

0

Think about the power supplied to the bulb. Assume for a moment constant voltage source, and constant resistance for each bulb (not true for bulb but often used to simplify discussion at this level) then in series you have a total resistance of $2R$ and power $P= VI = \frac{V^2}{2R}$ . This power is split by two bulbs so each sees $V^2/4R$. When the bulbs ...

0

The bulbs will only appear brighter if the available current to the system is not limited. In that case the series bulbs will have a lower voltage across each individual bulb and they will appear dimmer. If the power input to the circuit is a constant than the total wattage output from all bulbs is also constant and the bulbs will all appear the same ...

0

Depends on the angular velocity of its rotation. Electromagnetic interactions propagates at light speed. This means, if the coil is rotating in a frequency low enough, and if the induction circuit is close enough, the magnetic fields will arrive in the other circuit, almost "statically". Therefore, as good approximation in this case, you can consider the ...

3

The torque on electricity generators is continuously adjusted to keep them running at a constant speed (e.g. 60Hz in the US and 50Hz in the UK). When you turn on some electrical item the current it consumes places a greater load on some electricity generator somewhere and this reduces the speed. To counter this, at the generating station more torque is ...

2

Remember, $\vec{\nabla} \times \vec{B} = \mu_0 \vec{J}$, and $\vec{J}$ will be zero anywhere except on the loop itself, where it will be singular. Do you mean, perhaps, that the line integral around the loop equals the current, a la Ampere's law?

2

Power generation and supply management is not easy and it is to their credit that most of the time power companies supply people with AC at the same voltage no matter what the demand for power is. So when we turn on appliances we do not see the voltage drop as a result. Or more realistically when everyone gets home from work and starts cooking/ boiling ...

0

Yes..there will be induced current since when you rotate the coil the electrons in the coil also rotate. and by Lorentz force there will be an induced current. May be small but there will be an current.Faraday's paradox

0

In general, a inductive 'loop' for picking up a magnetic field (as shown connected to the galvanometer in you diagram) is 'symmetric' with respect to rotation about the axis parallel to the magnetic field from the coil connected to the battery. If the 'loop' is square (as shown), the symmetry is perfect for 90 degree rotations. There may be some small ...

1

Your instincts are spot on. While it’s still common for people to refer to electricity and magnetism as different phenomena, they’ve been formally unified since Maxwell’s 1873 paper on the subject, and they were known to be intimately related for decades before that through Faraday’s work among others. “Electromagnetism” covers all of the behavior of ...

0

All magnetism originates from movement of charge not movement of current as you mention. Whether it is arising from flow of charge positive or negative ( current ). Spin alignment's as in a ferromagnetic. Spin is again charge in motion that is rotation.

1

Each LED has both a voltage rating and a current rating (100mA). When connected in parallel, they will all receive the voltage of your power supply and draw 100mA each, so the total will be 300mA. When connected in series, they will split the supply voltage between them, so the supply voltage will have to be 3 times the individual LED voltage rating. Each ...

1

The first equation is when you want to solve for either the voltage, current, or power already knowing the other two, similar for $P=I^2R$, when you want to solve for an unknown already having knowledge of the other two. The last equation you get by noting that, $P=IV=I^2R$, hence $V=IR$ or $I=\frac{V}{R}$ plugging this into, the first equation you get ...

Top 50 recent answers are included