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Circuit diagram

With reference to the above image:
Here the battery produces a constant (will it be constant? pl. explain) electric field across the length of the wire, lets say its value is \$\vec E\$ . This field exerts a force on the electrons in the wire and they start moving from higher potential to lower potential.

First of all what will be the value of this constant electric field (if it is so?). Will it be \$ |E| = \frac{\Delta V}{\Delta r}\$ where \$\Delta r\$ is the length of the whole wire across the circuit (including resistances)?

Now I have been told that the potential of +ve terminal of the battery and point A (in image) is same which concludes that there is no potential difference between +ve terminal and A thus there should be no electric field ! But electric field exists how? also if electric field exists then why there is no potential drop ?!!

Coming to resistances $R_1$ and \$R_2\$. Now this constant electric field enters $R_1$ and there is a drop in potential this time i.e. potential of point B is lesser than that of A. Now does the electric field across this resistor be same as calculated above i.e. \$ |E| = \frac{\Delta V}{\Delta r}\$ or will it change ? Here Ohm's law states the \$\Delta V_1 = iR_1\$ so does it conclude that electric field inside this resistor will be \$|E| = \frac{\Delta V_1}{\Delta r_1}\$ where \$\Delta r_1\$ is the length of the resistor \$R_1\$ . What will happen and which is true ??
What will be the electric field through resistor \$R_2\$ ? and what will be the filed between C and -ve termianl of the battery ?

Please explain all these ?

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Here the battery produces a constant (will it be constant? pl. explain) electric field across the length of the wire,

The electric field will be constant with respect to time, but it will not be uniform (with respect to location), throughout the circuit. The electric field obeys the microscopic version of Ohm's law. $$\vec{E} = \frac{\vec{j}}{\sigma}$$

where \$\vec{j}\$ is the current density and \$\sigma\$ is the conductivity. Only if the ratio of these two terms is uniform, for example in a uniform wire, will the electric field be uniform. This is unlikely in a circuit with lumped components.

Now I have been told that the potential of +ve terminal of the battery and point A (in image) is same

This is a very good approximation. The voltage drop in the wire between the battery and resistor will be small. But it is not exactly true. The voltage drop will be non-zero. It depends on the resistance of the wire and the current flowing through it.

so does it conclude that electric field inside this resistor will be \$|E| = \frac{\Delta V_1}{\Delta r_1}\$ where \$\Delta r_1\$ is the length of the resistor \$R_1\$ .

No, the electric field will follow the microscopic version of Ohm's law given above.

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