Distribution of Currents in a Network of Resistanceless Conductors
Airfare Daily Deals eCigarettes Eyeglasses Hotels Jewelry Online Backup Online Dating Online Printing Online Tickets Skin Care Textbook Rentals Vitamins Web Hosting Weddings
Find thousands of shopping-related forums
SEARCH

Distribution of Currents in a Network of Resistanceless Conductors

The analysis of current distribution in a circuit of normal conductors is done by help of Ohm’s law, Kirchhoff’s laws and other network theorems but trivial use of these laws in case of a resistanceless circuit lead to wrong results. For the case of normal conductors currents are mainly determined by the resistances but in a resistanceless circuit we need to consider self-inductances and mutual inductances also. This article describes the analysis of such non-trivial cases.

Analyzing how the total current will distribute in a network of resistors is a basic problem of Electronics and is done by using some laws and standard techniques. Some of the widely used laws are Ohm’s law, Kirchhoff’s voltage law and Kirchhoff’s current law. But these laws do not work if we make the conductors resistanceless, which is often the case of superconductors. In this article I will explain how to analyze the distribution of currents in a Network of resistanceless conductors.

To begin with, we will consider the case of normal conductors and then turn of their resistances to zero and see how the resistancelessness affects the current distribution. Let us consider the following circuit consisting of two resistors ACB and ADB connected to an external battery. The two resistors have resistances R1 and R2 respectively and the battery has an internal resistance r. We need to find what the amounts of currents I1 and I2 flowing through the resistors are. Here total current through the circuit I = I1 + I2 = V / ( R1||R2 + r ).

Now, the analysis is very simple and straight forward. Since the potential difference between the points A and B is same along the two paths ACB and ADB, using Ohm’s law we have,

     I1*R1 = I2*R2

or, I1/I2 = R2/R1

This implies that current is inversely proportional to the resistance of the resistor. Note that, here we have neglected the inductance of the resistors which is justified for normal resistors (since the effect of inductance is negligibly small). Now, if we reduce the resistance of the conductors to zero, how will the current I divide between branches ACB and ADB? Clearly Kirchhoff’s laws are of no help suggesting that equal currents will flow through each branch. But this not the actual case and now self-inductances of the conductors start to contribute. We shall now show that division of the current I is determined by those inductances.

Let L1 and L2 be the self-inductances of the conductors ACB and ADB and M be the mutual inductance between them. Now, keeping in mind that inductance times the rate of change of current gives the potential drop across an element, we equate the voltage difference between A and B to get,

In most cases magnetic coupling between two parallel conductors is so small that their mutual inductance can be neglected. Under such circumstance we can say that the currents in parallel paths are inversely proportional to the self-inductances of those conductors in case of a resistanceless network. Note that, here the total current I is different from the case of normal conductors and is equal to V/r.

Thus we have learnt how the current distributes itself in a network of resistanceless conductors.

Need an answer?
Get insightful answers from community-recommended
experts
in Physics on Knoji.
Would you recommend this author as an expert in Physics?
You have 0 recommendations remaining to grant today.
Comments (0)
ARTICLE DETAILS
RELATED ARTICLES
ARTICLE KEYWORDS