To measure the internal resistance of a battery, a potentiometer is used. For , the balance point is observed at = 500 cm and for the balance point is observed at = 400 cm. The internal resistance of the battery is approximately:

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
To measure the internal resistance of a battery, a potentiometer is used. For $R = 10 Ω$ , the balance point is observed at $l$ = 500 cm and for $R = 1 Ω$ the balance point is observed at $l$ = 400 cm. The internal resistance of the battery is approximately: [2024]

1 $0.2 Ω$

2 $0.4 Ω$

3 $0.1 Ω$

4 $0.3 Ω$

Ans.
(4)

At balance point, the current in circuit is $i = \frac{ε}{r + R}$

$V_{A B} = ε - i r = ε - (\frac{ε}{r + R}) \cdot r$

$V_{A B} = \frac{ε r + ε R - ε r}{r + R} = \frac{ε R}{r + R}$

also, $V_{A B} = λ ℓ, λ$ is potential gradient.

$∴ λ ℓ = \frac{ε R}{r + R}$

Now, $\frac{l_{1}}{l_{2}} = \frac{R_{1}}{r + R_{1}} \times \frac{r + R_{2}}{R_{2}} \Rightarrow \frac{500}{400} = \frac{10}{r + 10} \times \frac{r + 1}{1}$

$5 (r + 10) = 40 (r + 1) \Rightarrow (r + 10) = 8 (r + 1)$

$7 r = 2 \Rightarrow r = \frac{2}{7} \approx 0.3$

Want Us to Email you About Special Offers & Updates?