# Tag Info

0

You first need to know that how the current flows through a capacitor. There are two parallel plates with a dielectric between them such as air etc. One plate is connected to the positive terminal of the battery while other is connected to the negative terminal of the battery. Charges from the negative terminal will accumulate on the plate connected to it. ...

0

Suppose the e.m.f. is denoted by $V_0$, and the current in the circuit is $I$. Assuming a sign convention such that $I > 0$ when the current is clockwise, the first thing is to write $I$ in terms of $V_0$, $R$ and $r$. In the manner you have drawn the circuit, $r$ is the "internal resistance" of the "battery" whose e.m.f. is $V_0$. Remember, the e.m.f. is ...

3

The fact that the cell has internal resistance & it acts as a source corrupts your argument. Your argument that connecting a resistance across one of the cells in parallel reduces the overall resistance of the circuit and hence expect the current to increase is wrong. I will derive an equation to obtain the current & potential drop across the bulb ...

2

$1~\textrm{Ampere-hour}$ equals $3600~\textrm{Coulomb}$ of charge. The amount of charge a battery can hold is determined by the amount of chemicals that are in it and it's fixed by the design of the battery. Usually it only decreases because of loss of electrolyte or changes in the chemical and physical structure of the electrodes. The voltage on a ...

0

The 6V in 6V battery is a label which gives an indication of the sort of voltage which might be obtained from such a battery. For example if your ^ V battery was an lead-acid battery and it was fairly new and fully charged its voltage would be 6.3 volt and if older or partially discharges then it is more likely to be 6 V. A 1.5 V alkaline battery at the ...

0

EMF or Terminal voltage will be considered same if battery has no internal resistance. If battery has some internal resistance then terminal voltage will be different(less) from the EMF or potential difference from the battery. If a battery has internal resistance then what will be considered if the same statement is given.

0

In your questions all the cases are assumed to be ideal unless mentioned. Therefore the electromotive force and the terminal voltage are equal in that case as internal resistance of the battery is considered negligible(if not given) . If the battery has internal resistance then the emf remains constant,but the terminal voltage decrease by a value which is ...

5

As drawn, the circuit, assuming ideal circuit elements, is problematic for the reason you've deduced (KVL gives a contradiction). One interpretation is that there is infinite large current for an infinitesimal time which instantaneously charges the capacitors to their final steady state voltages. To gain some insight, add a resistance $r$ in series with ...

3

No, that circuit cannot exist in that regime. You are neglecting the internal resistance of the wires between the voltage source and the capacitors, and if the capacitors are discharged (in which case the voltage over them is zero) that's no longer a good approximation. You therefore need to insert a small resistance on either side of the voltage source, ...

2

Current only flows if there is a voltage across the resistors. If the resistors are ideally short circuited, no current can flow through them.

-1

Current would naturally not prefer to take a path with greater opposition. In such situations, the ratio of current passing through each wire is in the inverse ratio of its resistance. Apply kirchoff law and verify for yourself.

1

If you add more resistors in series the effect will be the opposite of what you say: the battery will last longer. A battery has a certain rated capacity, written in mAh (milliamps times hours). Divide this capacity by the current you are drawing and you will get how much will that battery last in hours, at the same current draw. More resistance means less ...

0

Within a battery a chemical reaction is responsible for moving mobile electrons from one terminal to the other. The terminal from which the mobile electrons come from is called the positive terminal (deficit of mobile electrons) of the battery and the one that they go to is called the negative terminal (surplus of negative electrons). If there is no ...

0

Suppose that you double the separation of the plates of a parallel plate capacitor of initial capacitance $C$ which is connected to a battery of emf $V$. The capacitance of the capacitor become $\frac C 2$ and the energy stored in the capacitor changes from $\frac 12 CV^2$ to $\frac 1 2 \frac C 2 V^2$. There is a decrease in the energy stored in the ...

0

I had thought that for part A both light bulbs would begin glowing since the capacitor isn't charged, but i have no idea how to tell which one is brighter. I would reason about this in the following way. (1) This is a series circuit and so, the current through each circuit element (battery, bulbs, capacitor, switch) is identical. (2) The bulbs are ...

1

You can solve this problem by using Kirchhoff's two laws. Kirchooff's current law tells you that since this is a series circuit the current in each part of the circuit will be the same all the time. So whatever happens to bulb $A$ will also happen to bulb $B$ as they both have the same current flowing through them. Using Kirchhoff's voltage law you have ...

1

You're on the right track. You're probably familiar with how the current decreases exponentially after closing the switch. $$I(t)=I_0 e^{-t/\tau}$$ Where $\tau$ is the time constant of the circuit given by $\tau = RC$, and $R$ is the total resistance of the bulbs. So when the switch is closed, current will be a maximum, and the bulbs brightest. As time ...

4

a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...

-2

You have to keep this important thing in mind while dealing with problems of these kind. A capacitor behaves as a pure conductor immediately after connecting it with the circuit (i.e., at time t=0). As time passes the capacitor begins to lose its conductance exponentially and finally after a very long time (theoretically when time t tends to infinity) it ...

2

I would say it is both! Because of the abundance of electrons, the electric field at the battery pole/boundary, at the instant of turning on the switch (t=t0), is quickly (within a few Debye lengths) screened and cannot possibly reach the electrons further down the wire. However, the electrons at the vicinity of the pole that do feel the effect of electric ...

1

I want to charge a 12v 100ah battery which need 20 amp of current. You can charge the battery at any current proving it is not too high and it might be that the 20 amp is the maximum charging current? If the adapter gives a constant (regulated) 12 volts then you will not be able to charge a battery of the lead-acid type it will require more than 12 V ...

Top 50 recent answers are included