Question

In a potentiometer experiment, when three cells $A, B$ and $C$ are connected in series the balancing length is found to be $740\, cm$. If $A$ and $B$ are connected in series balancing length is $440\, cm$ and for $B$ and $C$ connected in series that is $540 \,cm$. Then the emf of ${{E}_{A}},{{E}_{B}}$ and ${{E}_{C}}$ are respectively (in volts)

• A. 1, 1.2 and 1.5
• B. 1, 2 and 3
• C. 1.5, 2 and 3
• D. 1.5, 2.5 and 3.5

Answer 1

Answer: (A)

1, 1.2 and 1.5

Answer 2

${{V}_{A}}+{{V}_{B}}+{{V}_{C}}\propto 740$
${{V}_{A}}+{{V}_{B}}\propto 440$
${{V}_{B}}+{{V}_{C}}\propto 540$
Hence, ${{V}_{A}}:{{V}_{B}}:{{V}_{C}}=1:1.2:1.5$