Question

The emf of the cell is 2V and its internal resistance is 2 ohm. A resistance of 8 ohm is joined to the battery in parallel. This is connected is the secondary circuit of the potentiometer. If 1V standard cell balances for 100 cm of potentiometer wire, the balance point of the above cell is :

• A. $120cm$
• B. $240cm$
• C. $160cm$
• D. $116cm$

Answer 1

The correct option is C $160cm$

For $1V,1=k\left(100\right).....\left(1\right)$ where $K$ is the potential gradient

$k=\frac{1}{100}V/em$....

After connecting $8\Omega$ parallel to $2V$ battery , [balanced condition ]

$i=\frac{E}{r+R}=\frac{2}{2+8}=0.2A$

$V_A-2i+2=V_B,V_A-V_B=2i-2=2\times 0.2-2$

$=-1.6V$

$\therefore V_B-V_A=1.6V$

$V_B-V_A=k{l}_2$....... $\left(2\right)$

Dividing equation $\left(1\right)$ by $\left(2\right)$,

$\frac{1}{V_B-V_A}=\frac{100}{l_2},\frac{1}{1.6}=\frac{100}{l_2}$

$l_2=160cm$