N number of cells , each of emf E and internal resistance r are connected in series so as to form a closed loop . The potential difference between the two terminals of each cell is ?
Answers
Answer:=2E/n
Explanation:Number of cells= n
EMF of each cell= E
and internal resistance of each cell= r
Total resistance in circuit= nr
One cell is joined in reverse, so total emf = nE-2E
Current in the circuit, I= nE-2E /nr
=E(n-2)/nr
Potential difference ,V = E- Ir
= E-(n-2)E×r /n×r
= E[1-(n-2)/n]
=2E/n
Explanation:
Assuming that by “PD” you mean “Potential Difference” or voltage, this is how I’d solve the problem:
Let’s assume as you said that the (open circuit) voltage across each “cell” (by which I’ll further assume you mean a battery comprised of one or more cells) is E, and that there are N of these batteries, each with its own internal resistance R. If you connect all the batteries in series correctly (with the positive terminal of each battery connected to the negative terminal on the next), the total open-circuit voltage of the string is
E (total = N٠E. The total series resistance of this string R(total) = N٠R.
You didn’t specify a load resistance, so I’m assuming the loop current in your circuit is limited only by the internal resistance of the batteries. That means that the loop current in the first case is I = (N٠E)/(N٠R) = E/R
If you connect one battery backwards, the total string voltage is
E (total) = (N-2)٠E
Note that the 2 is because if we assume the batteries are all identical, the miswired battery will cancel out the voltage of one of the correctly-wired batteries in the string.The total series resistance of the string remains the same, R(total = N٠R.
The loop current in this configuration is I = (N-2)٠E / N٠R.
The voltage across the miswired cell is equal to the IR (I٠R) drop across its internal resistance, plus the voltage across the cell (recall that under load, a battery’s terminal/output voltage is equal to its open circuit voltage minus the IR drop across its internal resistance).
Thus in the miswired case, E(cell) = [(N-2)٠E / N٠R] + E = E [(N-2)/N٠R) + 1] .
In the special case where there are only two batteries (N = 2), one of which is wired backwards, then the net loop voltage is zero, and the loop current is zero. In this case only, the voltage across the miswired battery will be equal to its open circuit voltage E.