You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Electric Bulbs

A potentiomet...

Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2 V$ and a negligible internal resistance. The potentiometer wire itself is $4 m$ long. When the resistance $R$ is connected across the given cell has value of $(i)$ infinity, $(i i) 9.5 Ω$ the balancing length on the potentiometer wire are found to be $3 m$ and $2.85 m$ respectively. The value of internal resistance of the cell is

0.25 Ω

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.95 Ω

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 Ω

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.75 Ω

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.5 Ω$
Given,
$l_{1} = 3 m,$
$l_{2} = 2.85 m$ ,
$R = 9.5 Ω$

Using, $r = (\frac{l_{1}}{l_{2}} - 1) \times R$
$r = (\frac{3}{2.85} - 1) \times 9.5$
$∴ r = \frac{0.15}{2.85} \times 9.5 = 0.5 Ω$

Suggest Corrections

Similar questions

Q. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of

2 V

and a negligible internal resistance. The potentiometer wire is

4 m

long. When the resistance

R = 9.5 Ω

is connected across a secondary cell the balancing length on the potentiometer wire was found to be

2.85 m

. If the balancing length for the emf of the secondary cell is

3 m

, the value of internal resistance of the cell is

Q. A

10 m

long uniform metallic write having a resistance of

20 Ω

is used as a potentiometer wire. This wire is connected in series with another resistance of

480 Ω

and a battery of emf

5 V

having negligible internal resistance. If an unknown emf e is balanced across

6 m

of the potentiometer wire, calculate:
(i) The potential gradient across the potentiometer wire.
(ii) The value of the unknown emf e.

Q. A potentiometer wire of length

100 cm

has a resistance of

10 Ω

. It is connected in series with a resistance and a cell of emf

2 V

and of negligible internal resistance. A source of emf

10 mV

is balanced against a length of

40 cm

of the potentiometer wire. What is the value of external resistance?

A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:

Q. In the potentiometer arrangement shown, the driving cell D has emf E and internal resistance r. The cell C, whose emf is to be measured, has emf E/2 and internal resistance 2r. The potentiometer wire is 100 cm long. If balance is obtained, the length

A J = l

Join BYJU'S Learning Program

The correct option is C 0.5 ΩGiven, l1=3 m, l2=2.85 m, R=9.5 Ω Using, r=(l1l2−1)×R r=(32.85−1)×9.5 ∴r=0.152.85×9.5=0.5Ω

The correct option is C $0.5 Ω$
Given,
$l_{1} = 3 m,$
$l_{2} = 2.85 m$ ,
$R = 9.5 Ω$

Using, $r = (\frac{l_{1}}{l_{2}} - 1) \times R$
$r = (\frac{3}{2.85} - 1) \times 9.5$
$∴ r = \frac{0.15}{2.85} \times 9.5 = 0.5 Ω$