Question

In a potentiometer wire experiment, the emf of a battery in the primary circuit is $20V$ and its internal resistance is $5\Omega$. There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from $120\Omega$ to $170\Omega$. Resistance of the potentiometer wire is $75\Omega$. The following potential differences can be measured using this potentiometer

• A. $5V$
• B. $6V$
• C. $7V$
• D. $8V$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $5V$
B $6V$
C $7V$

Here, $E=20$V$,r=5$ ohm, $R_l=75$ ohm and $R$ varies from 120 ohm to 170 ohm.
Now, Current when $R=120$ ohm is

$I=\frac{E}{R+r+R_l}$

$I=\frac{E}{120+5+75}$

$I=0.05$ A
Hence the voltage that can be measured is $V=0.05\times 120=6$ V
and when $R=170$ ohm, the current is $I=\frac{20}{170+5+75}=0.08A$

Hence the voltage that can be measured is $V=0.08\times 170=13.6$ V
Hence potential difference from 6V to 13.6 V can be measured using potentiometer.