Why is current same throughout a circuit

They also use less wiring than parallel circuits. The current is the same everywhere in a series circuit. It does not matter where you put the ammeter, it will give you the same reading. The current in a series circuit depends upon the number of cells.

If you make the cells face in the same direction, the more cells you add, the greater the current. If you put more lamps into a series circuit, the lamps will be dimmer than before because less current will flow through them. You might think that the current gets less as it flows through one component after another, but it is not like this. Charge does NOT become used up by resistors such that there is less of it at one location compared to another.

The charges can be thought of as marching together through the wires of an electric circuit, everywhere marching at the same rate. Current - the rate at which charge flows - is everywhere the same. It is the same at the first resistor as it is at the last resistor as it is in the battery.

Mathematically, one might write. These current values are easily calculated if the battery voltage is known and the individual resistance values are known. Using the individual resistor values and the equation above, the equivalent resistance can be calculated.

As discussed in Lesson 1 , the electrochemical cell of a circuit supplies energy to the charge to move it through the cell and to establish an electric potential difference across the two ends of the external circuit.

This is to say that the electric potential at the positive terminal is 1. As charge moves through the external circuit, it encounters a loss of 1. This loss in electric potential is referred to as a voltage drop. It occurs as the electrical energy of the charge is transformed to other forms of energy thermal, light, mechanical, etc. If an electric circuit powered by a 1.

There is a voltage drop for each resistor, but the sum of these voltage drops is 1. This concept can be expressed mathematically by the following equation:. To illustrate this mathematical principle in action, consider the two circuits shown below in Diagrams A and B. Suppose that you were to asked to determine the two unknown values of the electric potential difference across the light bulbs in each circuit.

To determine their values, you would have to use the equation above. The battery is depicted by its customary schematic symbol and its voltage is listed next to it. Determine the voltage drop for the two light bulbs and then click the Check Answers button to see if you are correct. Earlier in Lesson 1, the use of an electric potential diagram was discussed. An electric potential diagram is a conceptual tool for representing the electric potential difference between several points on an electric circuit.

Consider the circuit diagram below and its corresponding electric potential diagram. The circuit shown in the diagram above is powered by a volt energy source. There are three resistors in the circuit connected in series, each having its own voltage drop. The negative sign for the electric potential difference simply denotes that there is a loss in electric potential when passing through the resistor.

Conventional current is directed through the external circuit from the positive terminal to the negative terminal. Since the schematic symbol for a voltage source uses a long bar to represent the positive terminal, location A in the diagram is at the positive terminal or the high potential terminal.

Location A is at 12 volts of electric potential and location H the negative terminal is at 0 volts. In passing through the battery, the charge gains 12 volts of electric potential. And in passing through the external circuit, the charge loses 12 volts of electric potential as depicted by the electric potential diagram shown to the right of the schematic diagram.

This 12 volts of electric potential is lost in three steps with each step corresponding to the flow through a resistor. In passing through the connecting wires between resistors, there is little loss in electric potential due to the fact that a wire offers relatively little resistance to the flow of charge.

Since locations A and B are separated by a wire, they are at virtually the same electric potential of 12 V. When a charge passes through its first resistor, it loses 3 V of electric potential and drops down to 9 V at location C.

Since location D is separated from location C by a mere wire, it is at virtually the same 9 V electric potential as C. When a charge passes through its second resistor, it loses 7 V of electric potential and drops down to 2 V at location E. Since location F is separated from location E by a mere wire, it is at virtually the same 2 V electric potential as E.

Finally, as a charge passes through its last resistor, it loses 2 V of electric potential and drops down to 0 V at G. At locations G and H, the charge is out of energy and needs an energy boost in order to traverse the external circuit again. The energy boost is provided by the battery as the charge is moved from H to A. The Ohm's law equation can be used for any individual resistor in a series circuit.

When combining Ohm's law with some of the principles already discussed on this page, a big idea emerges. Wherever the resistance is greatest, the voltage drop will be greatest about that resistor. The Ohm's law equation can be used to not only predict that resistor in a series circuit will have the greatest voltage drop, it can also be used to calculate the actual voltage drop values.

The above principles and formulae can be used to analyze a series circuit and determine the values of the current at and electric potential difference across each of the resistors in a series circuit. Their use will be demonstrated by the mathematical analysis of the circuit shown below.

The goal is to use the formulae to determine the equivalent resistance of the circuit R eq , the current at the battery I tot , and the voltage drops and current for each of the three resistors. The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit.

Now that the equivalent resistance is known, the current at the battery can be determined using the Ohm's law equation. Connect and share knowledge within a single location that is structured and easy to search. So I am a 10th grade student and my teacher told me that the current is the same at every point in a series circuit. It does split up in parallel circuit but it then recombines and the current flowing out of the battery is the same as the current flowing back into it. I mean let's say that there is a light bulb somewhere in a series circuit.

Now, current or electrical energy will flow into it and then convert into light energy. Which is not possible. I'm surprised that no one has yet mentioned the hydraulic analogy for electricity to help the OP understand better. A brief summary of this analogy is:. Any point in a series circuit pipe has the same amount of water flowing past that point as any other. Any split in a pipe parallel circuit shares all water flowing into the split the current of all legs equals the current before the split.

There are also water analogies for other electrical components like coils and capacitors; visit the link if you are interested. Imagine a block sliding down a slope and that there is an amount of friction between the slope and the block such that the block slides down the slope at constant speed.

As the block slides down the slope it loses gravitational potential energy and an equal amount of heat is generated due to the friction between the slope and block. The block does not change but is part of the mechanism by which gravitational potential energy is converted into heat.

You can think of the electrons as being charged blocks which are losing electrical potential energy sliding along a potential gradient at the drift velocity and this energy is converted into heat and light in a light bulb. When the electrons reach the positive terminal of the voltage source they are then given more electrical potential energy by the voltage source and so can again slide along the potential gradient.

The electrons do not change and are not created or destroyed rather they are part of the mechanism by which electrical potential energy is converted into heat and light. Interesting question. As you could see from the examples with falling water or sliding blocks the gravitational potential is responsible for energy release in mechanical storage systems. For a battery this can't be the reason of energy storage. Kinetic energy inside the a battery can't be the reason too. At the end in the battery an electromagnetic interaction re-arrange the electrons into different chemical bonds.

And pushing electrons into re-arrangements some part of the EM radiation over goes to the electrons. Electrons are able to store EM radiation or more precise to store photons. Mostly some part of the energy will be released again in the form if longer wavelength photons and this is why the storage of energy in accumulators is inefficient.

So beside the potential gravitational energy and the kinetic energy there is a storage of EM energy. You feel it right with our question. The electrons at the end of the circuit are of the same amount and of course of the same charge. The release of energy in a bulb as well as in the wire due to the Ohm resistance is only possible because before the electrons are charged with EM energy.

Note, that such an explanation, if you could not find in literature, should not be wrong but could be new. So your question touches an interesting point of the interaction between electrons and EM radiation. Batteries have energy stored inside. The energy inside a battery could accelerate electrons to a very high velocity.

However, this doesn't happen in a circuit. Just as the battery speeds up an electron, the electron hits to another one and remains constant speed. So the battery constantly pushes the electron but in the end it comes back to it with the same velocity. The battery is constantly pushing but the electron remains in the same velocity, now it makes sense.

The energy released by the battery gets used in the circuit and becomes heat which can later become light.