Question

If a cell of internal resistance $1.5\Omega$ and emf of $1.5V$ balances $500cm$ on a potentiometer wire, now by what length would the balance point shift if a cell of internal resistance $2\Omega$ and emf of $2V$ is connected in parallel (along the same polarity) with the prior cell?

• A. $71.4cm$ towards left
• B. $71.4cm$ towards right
• C. $125cm$ towards right
• D. Does not shift

Answer 1

The correct option is B $71.4cm$ towards right

Case-1:
Case-2:

using kirchhoff's current law:
$i=\frac{E}{r_{equi}}=\frac{E_1}{r_1}+\frac{E_2}{r_2}=1+1=2$
total resistance= $\frac{2\times 1.5}{2+1.5}=\frac{6}{7}$

So, the EMF of the circuit,
$\frac{7E}{6}=2E=\frac{12}{7}$
Here, extra emf is adding for balancing, hence meter point will move towards right.

for the case-1
$1.5\to 500cm$
$\frac{12}{7}\to \frac{12/7\times 500}{1.5}=571.4cm$

Now, the shift of the meter $\Delta x=571.4-500=71.4cm$

so, option B is correct.