Heating Effect of Current and Electrical Measuring Instruments

Q 1 :

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]

$0.2 Ω$
$0.4 Ω$
$0.1 Ω$
$0.3 Ω$

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

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

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

Q 2 :

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_{p} = 1 Ω$ as shown in the figure. An external resistance of $R_{e} = 2 Ω$ is connected via the sliding contact. [2025]

0.3 A
1.35 A
1.0 A
0.9 A

(3)

The circuit can be considered as

$∴$ $R_{e q} = 0.5 + \frac{0.5 \times 2}{2 + 0.5} = (\frac{5}{10} + \frac{10}{25}) Ω$

$= \frac{45}{50} = \frac{9}{10} = 0.9$

$∴$ $i = \frac{0.9}{0.9} = 1 A$

Topic Question Set

Q 1 :

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]

Q 2 :

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_{p} = 1 Ω$ as shown in the figure. An external resistance of $R_{e} = 2 Ω$ is connected via the sliding contact. [2025]

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