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]

$0.2 Ω$
$0.4 Ω$
$0.1 Ω$
$0.3 Ω$

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2

3

4

(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

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2

3

4

(3)

The circuit can be considered as

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

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

$∴$ $i = \frac{0.9}{0.9} = 1 A$

Q 3 :

With the help of a potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is [2023]

(A) directly proportional to the length of the potentiometer wire

(B) directly proportional to the potential gradient of the wire

(C) inversely proportional to the potential gradient of the wire

(D) inversely proportional to the length of the potentiometer wire

Choose the correct option for the above statements:

B and D only
A and C only
A only
C only

1

2

3

4

(2)

Sensitivity of potentiometer wire is inversely proportional to potential gradient.

Q 4 :

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 $Ω$ . When a resistance of 15 $Ω$ is used for shunting null point moves to 300 cm. The internal resistance of the cell is ______ $Ω$ . [2023]

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(5)

$\frac{ε}{r + 5} \times 5 = 200 x$ ....(i)

$\frac{ε \times 15}{r + 15} = 300 x$ ...(ii)

$\Rightarrow r = 5 Ω$

Q 5 :

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length of null point increases by 40 cm. The value of E is $\frac{x}{10}$ V. The value of $x$ is ______. [2023]

0%

(25)

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\frac{1.5}{E_{2}} = \frac{60}{60 + 40} = \frac{6}{10} = \frac{3}{5}$

$E_{2} = \frac{5}{2} = \frac{x}{10} \Rightarrow x = 25$

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]

Q 4 :

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 $Ω$ . When a resistance of 15 $Ω$ is used for shunting null point moves to 300 cm. The internal resistance of the cell is ______ $Ω$ . [2023]

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