What is the potential drop between points A and C in the following circuit ?
Resistances $1 \, \Omega $ and $2 \, \Omega $ represent the internal resistances of the respective cells.
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Updated On: Aug 1, 2023
$1.75\,V$
$2.25\,V$
$ \frac{5}{4}{V} $
$ \frac{4}{5} V $
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The Correct Option isB
Approach Solution - 1
Real : Emf's $E_{1}$ and $E_{2}$ are opposing each other.
Since, $E_{2}>E_{1}$ current will move from right to left.
Current in circuit
$i=\frac{E_{2}-E_{1}}{R+r_{1}+r_{2}}$$=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25 A$
The potential drop between points $A$ and $C$ is
$V_{A}-V_{C} =E_{1}+i r_{1} $$=2+(0.25 \times 1) $$=2.25 \,V$
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Approach Solution -2
Emf's $E_{1}$ and $E_{2}$ are opposing each other.
Since, $E_{2}>E_{1}$ current will move from right to left.
Current in circuit
$i=\frac{E_{2}-E_{1}}{R+r_{1}+r_{2}}$$=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25 A$
The potential drop between points $A$ and $C$ is
$V_{A}-V_{C} =E_{1}+i r_{1} $$=2+(0.25 \times 1) $$=2.25 \,V$
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Approach Solution -3
Emf's $E_{1}$ and $E_{2}$ are opposing each other.
Since, $E_{2}>E_{1}$ current will move from right to left.
Current in circuit
$i=\frac{E_{2}-E_{1}}{R+r_{1}+r_{2}}$$=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25 A$
The potential drop between points $A$ and $C$ is
$V_{A}-V_{C} =E_{1}+i r_{1} $$=2+(0.25 \times 1) $$=2.25 \,V$
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0
0
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Approach Solution -4
Emf's $E_{1}$ and $E_{2}$ are opposing each other.
Since, $E_{2}>E_{1}$ current will move from right to left.
Current in circuit
$i=\frac{E_{2}-E_{1}}{R+r_{1}+r_{2}}$$=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25 A$
The potential drop between points $A$ and $C$ is
$V_{A}-V_{C} =E_{1}+i r_{1} $$=2+(0.25 \times 1) $$=2.25 \,V$
