Five identical cells each of internal resista | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $n = 5$ be the number of cells,$E = 5\, V$ be the $emf$ of each cell,and $r = 1\, \Omega$ be the internal resistance of each cell.For series c

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $n = 5$ be the number of cells,$E = 5\, V$ be the $emf$ of each cell,and $r = 1\, \Omega$ be the internal resistance of each cell.For series combination,the total $emf$ is $nE$ and total internal resistance is $nr$. The current $i_1$ is given by:$i_1 = \frac{nE}{R + nr} = \frac{5 	imes 5}{R + 5 	imes 1} = \frac{25}{R + 5}$For parallel combination,the total $emf$ is $E$ and total internal resistance is $r/n$. The current $i_2$ is given by:$i_2 = \frac{E}{R + r/n} = \frac{5}{R + 1/5} = \frac{5 	imes 5}{5R + 1} = \frac{25}{5R + 1}$Given $i_1 = i_2$,we have:$\frac{25}{R + 5} = \frac{25}{5R + 1}$$R + 5 = 5R + 1$$4 = 4R$$R = 1\, \Omega$

Five identical cells each of internal resistance $1\, \Omega$ and $emf$ $5\, V$ are connected in series and in parallel with an external resistance $R$. For what value of $R$ will the current in the series and parallel combination remain the same? (in $\Omega$)

Similar Questions

Ten identical cells each emf $2 \, V$ and internal resistance $1 \, \Omega$ are connected in series with two cells wrongly connected. $A$ resistor of $10 \, \Omega$ is connected to the combination. What is the current through the resistor (in $ \, A$)?

Write the equation for the equivalent electromotive force (emf) of cells connected in parallel.

In how many ways can a combination of cells be done? Explain.

$A$ cell of e.m.f. $E$ is connected with an external resistance $R$,then the potential difference across the cell is $V$. The internal resistance of the cell will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module