(B) The two cells are connected in series. The total $emf$ of the circuit is $E + E = 2E$.The total resistance of the circuit is $R + R_1 + R_2$.The c

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(B) The two cells are connected in series. The total $emf$ of the circuit is $E + E = 2E$.The total resistance of the circuit is $R + R_1 + R_2$.The current $i$ flowing through the circuit is given by $i = \frac{2E}{R + R_1 + R_2}$.The potential difference $V$ across the cell with internal resistance $R_2$ is given by $V = E - i R_2$.Given that $V = 0$,we have $0 = E - i R_2$,which implies $E = i R_2$.Substituting the value of $i$ into this equation:$E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$$1 = \frac{2 R_2}{R + R_1 + R_2}$$R + R_1 + R_2 = 2 R_2$$R = R_2 - R_1$.

Class 12 Physics Current Electricity Electric Cells and Combination of cells in Series and Parallel

Difficult

Two sources of equal $emf$ are connected to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

A
$R = \frac{R_2(R_1 + R_2)}{(R_2 - R_1)}$
B
$R = R_2 - R_1$
C
$R = \frac{R_1 R_2}{(R_1 + R_2)}$
D
$R = \frac{R_1 R_2}{(R_2 - R_1)}$

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Current Electricity

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Electric Cells and Combination of cells in Series and Parallel

Three identical cells,each of emf $10 \ V$ and internal resistance $3 \ \Omega$,are connected in parallel across terminals $A$ and $B$. The net emf across $A$ and $B$ is .......... $V$.

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For two cells having the same $EMF$ $E$ and internal resistance $r$,the current passing through an external resistor of $6 \ \Omega$ is the same when both cells are connected either in parallel or in series. The value of internal resistance $r$ is . . . . . . $ \ \Omega$.

DifficultJEE MAIN 2026

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$N$ identical cells are connected in series or in parallel. When an external resistor $R$ is connected to them,they provide the same current. The internal resistance $r$ of each cell is:

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In the network shown,the potential difference between $A$ and $B$ is ................. $V$ $(R = r_1 = r_2 = r_3 = 1 \Omega, E_1 = 3 \, V, E_2 = 2 \, V, E_3 = 1 \, V)$.

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Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. If one of the cells is connected wrongly,the equivalent emf and effective internal resistance of the combination are

DifficultKCET 2015

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Two sources of equal $emf$ are connected to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

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Three identical cells,each of emf $10 \ V$ and internal resistance $3 \ \Omega$,are connected in parallel across terminals $A$ and $B$. The net emf across $A$ and $B$ is .......... $V$.

For two cells having the same $EMF$ $E$ and internal resistance $r$,the current passing through an external resistor of $6 \ \Omega$ is the same when both cells are connected either in parallel or in series. The value of internal resistance $r$ is . . . . . . $ \ \Omega$.

$N$ identical cells are connected in series or in parallel. When an external resistor $R$ is connected to them,they provide the same current. The internal resistance $r$ of each cell is:

In the network shown,the potential difference between $A$ and $B$ is ................. $V$ $(R = r_1 = r_2 = r_3 = 1 \Omega, E_1 = 3 \, V, E_2 = 2 \, V, E_3 = 1 \, V)$.

Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. If one of the cells is connected wrongly,the equivalent emf and effective internal resistance of the combination are

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