SATHEE: Current Electricity - Result Question 50

Current Electricity - Result Question 50

53. A cell can be balanced against $110 c m$ and 100 $c m$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 Ω$ . Its internal resistance is

[2008]

(a) $1.0 o h m$

(b) $0.5 o h m$

(d) zero

Show Answer

Answer:

Correct Answer: 53. (a)

Solution:

(a) Here $E > \frac{E R}{R + r}$ , hence the lengths $110 c m$ and $100 c m$ are interchanged.

Without being short-circuited through R, only the battery E is balanced.

$E = \frac{V}{L} \times l_{1} = \frac{V}{L} \times 110 \dots . . (i)$

When $R$ is connected across E, $R i = \frac{V}{L} \times l_{2}$

or, $R (\frac{E}{R + r}) = \frac{V}{L} \times 100$ .

Dividing (i) by (ii), we get

$\frac{R + r}{R} = \frac{110}{100}$

or, $100 R + 100 r = 110 R$

or, $10 R = 100 r$

$∴ r = \frac{10 R}{100} = \frac{10 \times 10}{100} (∴ R = 10 Ω)$

$\Rightarrow r = 1 Ω$ .

Current Electricity - Result Question 50

53. A cell can be balanced against $110 c m$ and 100 $c m$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 Ω$ . Its internal resistance is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Current Electricity - Result Question 50

53. A cell can be balanced against 110cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

53. A cell can be balanced against $110 c m$ and 100 $c m$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 Ω$ . Its internal resistance is