Is current flowing into a battery always equal to current flowing out?

[laser]

Homework Statement

See image in description

Relevant Equations

V=iR

Image from Wiley Intro Physics Textbook sample problem

This was the given problem. In the diagram, one can see that the current flowing into each battery equals the current flowing out of each battery. Is this always the case, and why?

If there was only one battery, I guess it could be considered conservation of charge, but for eg the battery on the left, why can't 2A flow out and 1A flow in?

 

[laser]

"Concentrating charge in a small circuit element requires a great deal of energy. Because charge is conserved one must make a rarefaction of charge somewhere else thereby roughly doubling the required energy input. Not gonna spontaneously occur. Nature is fickle"
?

 

[hutchphd]

Concentrating charge in a small circuit element requires a great deal of energy. Because charge is conserved one must make a rarefaction of charge somewhere else thereby roughly doubling the required energy input. Not gonna spontaneously occur.
THIS WAS MY RESPONSE ORIGINALLY. Don't know what screwed up......but something did.

 

[kuruman]

It's in Maxwell's equations. See charge conservation.

 

[DaveE]

For practical engineering problems you can assume that the electrons stay in the schematic and don't accumulate anywhere (even in capacitors!). If they go in somewhere, they also have to come out essentially instantly. Each branch of the schematic has a single current value.

 

[SammyS]

Homework Statement: See image in description
Relevant Equations: V=iR

If there was only one battery, I guess it could be considered conservation of charge, but for eg the battery on the left, why can't 2A flow out and 1A flow in?

Suppose, as you propose, that for the battery on the left we have 2A flow out and 1A flow in.
After 1 second of time in this situation, this battery would have sent out 1 Coulomb more charge than what it had received. Conservation of charge should tell you that this battery would then have acquired a net charge of $-\,1$ Coulomb. Do you see a problem with that ?

 

[Steve4Physics]

Homework Statement: See image in description
Relevant Equations: V=iR

Image from Wiley Intro Physics Textbook sample problem
View attachment 342906
This was the given problem. In the diagram, one can see that the current flowing into each battery equals the current flowing out of each battery. Is this always the case, and why?

If there was only one battery, I guess it could be considered conservation of charge, but for eg the battery on the left, why can't 2A flow out and 1A flow in?

Hi @laser. In addition to what's already been said, I wonder if the ambiguous meaning of 'charge' is causing any confusion. Just in case...

A battery does not store charge. A battery has zero electrical (+ or -) charge.

In normal use, we say a battery ‘discharges’. That means it converts its chemical energy to electrical energy. It’s supply of chemical energy is reducing.

Current-in = current-out. Always. (Think of a pump, pumping water around a heating system’s circuit: rate of water in = rate of water out.)

When we ‘charge’ a (rechargeable) battery, we’re not giving it any additional charge. We’re running a current through the battery ’backwards’. This converts electrical energy back to chemical energy inside the battery. That’s what’s meant by ‘charging a battery’.

Whether ‘discharging’ or ‘charging’, current-in = current-out. Always. (As already explained in previous posts).

 

[kuruman]

If they go in somewhere, they also have to come out essentially instantly.

Not in a charging capacitor. That's why the charge continuity equation is needed. Enclose only the positive plate of the charging capacitor with a box of volume $V$ and surface area $S$ (see figure on the right). Starting with the differential form of the continuity equation, integrate over the volume, and use the divergence theorem: $$\begin{align} & \mathbf{\nabla}\cdot \mathbf J+\frac{\partial \rho}{\partial t}=0 \nonumber \\
& \int_V(\mathbf{\nabla}\cdot \mathbf J)dV=-\frac{\partial}{\partial t}\int_V\rho~dV \nonumber \\
& \int_S\mathbf{J}\cdot \mathbf{\hat n}~dA=-\frac{\partial}{\partial t}\int_V\rho~dV =-\frac{dq_{encl.}}{dt} \nonumber \\
\end{align}$$ The only place on the surface of the box where the current density $\mathbf J$ is non-zero is the area of point P where the conducting wire enters the volume. Since $\mathbf{\hat n}$ is the outward normal and $\mathbf J$ is directed into the volume, $$\int_S\mathbf{J}\cdot \mathbf{\hat n}~dA=-\int_SJ~dA=-I$$ Thus, $I=\dfrac{dq_{encl.}}{dt}.$ This says that the charge per unit time that goes into the closed surface accumulates at the same rate inside the surface. Charge goes in but doesn't come out unless the capacitor is discharged.

 

[DaveE]

Yep, I just knew a physicist would step in with Maxwell's Equations etc. OK you win.
But... not really helpful for someone learning at this level.
Schematics are used to convey that really useful, but unrealistic, approximations are being used. They should only rarely be mixed with Maxwell and E&M.
Fortunately Einstein and SR hasn't been mentioned yet.

BTW, that's not a capacitor you cite. It's half of a capacitor, at least for an EE.

 

[kuruman]

BTW, that's not a capacitor you cite. It's half of a capacitor, at least for an EE.

To a physicist the figure shows a complete capacitor with a Gaussian (imaginary) closed surface enclosing one of its plates.

But... not really helpful for someone learning at this level.

I would agree with that 100%. It suffices to say what has already been said. Charge conservation is a law of Nature. The amount of charge that accumulates at point A in the circuit must be equal to the same amount that leaves another point B in the circuit. The circuit starts out with zero net charge and remains so at all times.

 

[jbriggs444]

Coulomb's constant $k$, the constant in $F=k \frac{q_1 q_2}{r^2}$ is about $9 \times 10^9$ Newton meter² per coulomb²

If you have one circuit element that accumulates one coulomb of charge and is one centimeter away from another circuit element that has accumulated the matching minus one coulomb of charge then this means that the electrostatic attraction between the two would be $9 \times 10^{13}$ Newtons. That is roughly the weight of one billion metric tons.

If I am integrating properly in my head it would take about 83 thousand kilowatt hours to create that charge distribution. [Integrating $q^2 dq$ has a $\frac{1}{3}$ in there. So you multiply $9 \times 10^{13}$ Newtons by $10^{-2}$ meters and divide by three to get Joules. Then divide by 3.6 million to get kwh]

The last bit of incremental charge flowing will be moving against a potential difference that is not divided by three. I make it just shy of one teravolt in this scenario.

 

[DaveE]

To a physicist the figure shows a complete capacitor with a Gaussian (imaginary) closed surface enclosing one of its plates.

Yes, got it. Everything is a capacitor. Thanks for reminding us.

 

[Mister T]

If there was only one battery, I guess it could be considered conservation of charge, but for eg the battery on the left, why can't 2A flow out and 1A flow in?

Same reason. Batteries don't create charge.