0
$\begingroup$

My problem is that there are n cells connected in parallel, connected to an external resistor.

Now can we use Kirchoff's loop rule to evaluate the current through the resistor by considering a loop from the +ve terminal of a single cell to the -ve terminal of that cell?

If we do then wouldn't the equation be
$E-iR=0$
and $i$ will come out to be $E/R$? This implies that the current in resistor $R$ would remain constant even if a cell is removed. But superposition of current says it will be reduced.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ By "n cells" do you mean n identical voltage sources? $\endgroup$ – garyp Jul 23 '16 at 15:45
  • $\begingroup$ Yep n identical Voltage sources $\endgroup$ – Shrish Dutta Jul 23 '16 at 15:47
  • $\begingroup$ Understanding what you ask seems much harder as answering your question. But don't worry, simply give a lot of more info, maybe even your circuit diagram. $\endgroup$ – peterh Jul 23 '16 at 15:47
  • $\begingroup$ In that case you are not applying Kirchoff's rules correctly. Work it out in detail for two cells. $\endgroup$ – garyp Jul 23 '16 at 15:49
  • $\begingroup$ I added a picture, is my problem clear now? $\endgroup$ – Shrish Dutta Jul 23 '16 at 16:04
0
$\begingroup$

Based on your drawing, with the batteries all in parallel, the resistor will have the same voltage across it whether you have one battery in the circuit or more than one battery in the circuit. Assuming that each battery produces the same emf, this will give you one value of current through the resistor, regardless of how many batteries are in parallel. What will change is the amount of current supplied by each battery. For 4 batteries in parallel, the current through each battery will be 1/4 of what it would be if you had only one battery in the circuit.

$\endgroup$
0
$\begingroup$

Your application of the Loop Rule is correct. The current around this loop is $i=E/R$.

However, although this is the only current going through that particular cell, it is not the only current going through the common resistor $R$. There are $n-1$ other loops through $n-1$ identical cells, each with the same loop current $i$. These loop currents are superposed, and add up to give the total current.

So after superposing all the loop currents, the current through each cell is $i=E/R$ and the total current through the common resistor is $ni=nE/R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.