# Voltage and Electric Field in a Circuit

I have a doubt regarding how the electric field acts in a circuit.

I have been told that a normal cell creates a uniform electric field, but I find it a bit confusing. Let me explain my doubt with a diagram.

Suppose there is a circuit and the resistor $r_1$ undergoes a potential drop of $0.4~\rm V$ and resistor $r_2$ undergoes a potential drop of $1.1~\rm V$ but then, since the electric field increases with increase in potential gradient, then the electric field through the resistor $r_2$ should be greater, shouldn't it? This goes against the idea that electric field in a circuit is uniform and if the electric field isn't constant in the circuit, then why is the battery called a fixed voltage source if the electric field applied by it varies? I find this very confusing,

• What gave you the idea that electric field in a circuit is uniform? It definitely isn't. Perhaps you're confusing current with electric field? – Chris Feb 14 '18 at 3:55
• since the cell produces a constant electric field, i thought it would be reasonable to assume electric field in a circuit is constant and even in the derivation of ohm's law ,it is taken a constant . – Adithya Vijay Feb 14 '18 at 3:59
• I don't know what derivation of Ohm's law you're talking about, but I think it's likely that the electric field is taken to be constant across a single component. As for your other point, you've shown that the electric field is not uniform in this circuit. The electric field configuration created by a cell depends on the circuit it is connected to, and is not uniform in general. – Chris Feb 14 '18 at 4:02
• so then , both the resistors should be experiencing different electric field ,right ?but since, a 1.5v potential difference is created across the terminals of the cell,an electric field would be generated which takes the electrons from the postive to negative terminal of the cell . shouldn't that be constant ? if then ,the electric field through the wire ,changes on seeing the resistor and after passing out of the resistor,it gains its constant magnitude .is that how it works ? – Adithya Vijay Feb 14 '18 at 4:16
• also, why is battery called a fixed voltage source if electric field supplied by it varies across resistors ? – Adithya Vijay Feb 14 '18 at 4:24

It is best not to think in terms of fields if you are doing circuits. The primary quantities of interest in circuits are voltage and current. The E field is the spatial derivative of the voltage, eg the units of the E field is V/m. However, in a circuit the position is abstracted away. There is no indication of the length of any wire or resistor or battery. So there is no way to obtain information about the E fields given only information about the circuit theory representation of the circuit.

Inside a resistor the E field should be more or less uniform, with a value approximately equal to the voltage across it divided by its length.

Inside a capacitor the E field should be more or less uniform between the plates and zero within the plates.

Inside a battery the E field should be quite strong right at the electrode where the electrochemical reaction makes the voltage difference occur over molecular-scale distances. The E-field will be substantially lower elsewhere, including approximately 0 within the metallic conductor and relatively close to 0 within the electrolyte. Calling it uniform is not accurate.

• > "Inside a resistor the E field should be more or less uniform" Not for heavy duty resistors that are made of coiled high resistance wire. Only magnitude of electric field is more-or-less the same along that wire. – Ján Lalinský Apr 21 at 23:32

Let the resistances 'r1' and 'r2' be replaced by their equivalent resistance 'R'. Ok. So, what we have now is the following equation---> V(=1.5V)=R(=r1+r2)xI(=current through the circuit) So, according to the diagram we can easily conclude that in the equivalent circuit [where we used the equivalent resistance R(=r1+r2)]---> V=RI=(r1+r2)XI=(Ixr1)+(Ixr2)=V1+V2 (which is true for the diagram circuit) So, V=V1+V2 is always true for the above circuit. The battery creates an uniform electric field through the circuit which results in flowing of current through the circuit, when this current crosses resistances it creates a potential change which is only true for the junctions of the resistances, so this case is also true for the other resistances in the circuit, the total amount of potential drop of the circuit will be equal to the potential drop caused by the battery in an equivalent circuit. [My discussion is based on series combination of resistances in a circuit which is same as the above picture]. To conclude it can be easily said that not for the battery(which is a fixed voltage source) its the value of a resistances that determines the potential drop across it. Hope you understand your confusion.

As you have noted, electric field in a circuit in not uniform. This is totally correct. The thing which is uniform is current.

But the total potential drop across the battery is $$1.5V$$. Even if you add more resistances and make different arrangements from them, the potential drop across the battery remains same. Hence, the Electric field across the battery remains constant. For this reason, the battery is called a fixed voltage source, it maintains constant potential drop across itself.